algos-and-data-structs

[sharp junco]

(was thinking of commenting the uneeeded lines)

[glass swan]

Hi below is code which has puzzled me about dictionaries:
Love it, if anyone could explain why this behavior is for dictionary and not for other variables.

def count_char(text, count={}, num=0):
    for t in text:
        count[t] = 1 + count.get(t,0)
    num += 1
    print("dict - inside function =",count)
    print("var - inside function", num)
    return count

count_char('abc')
count_char('efg')
count_char('hij')

Output :

dict - inside function = {'a': 1, 'b': 1, 'c': 1}
var - inside function 1
dict - inside function = {'a': 1, 'b': 1, 'c': 1, 'e': 1, 'f': 1, 'g': 1}
var - inside function 1
dict - inside function = {'a': 1, 'b': 1, 'c': 1, 'e': 1, 'f': 1, 'g': 1, 'h': 1, 'i': 1, 'j': 1}
var - inside function 1

[flint arrow]

Why when i change text in file it still prints old information ?

# Using readline()
lines = [0,1,2,3,4,5,6,7]
i = 0
result = []
while True:
    with open("myfile.txt", "r+") as fp:
        # access each line
        for line in fp:
            # check line number
            if i in lines:
                result.append(line.strip())
            i = i + 1
            
    testas1 = result[0]
    testas2 = result[1]
    testas3 = result[2]
    testas4 = result[3]
    testas5 = result[4]
    testas6 = result[5]
    testas7 = result[6]
    testas8 = result[7]
    print(testas1)
    print(testas2)
    print(testas3)
    print(testas4)
    print(testas5)
    print(testas6)
    print(testas7)

[slender sandal]

It happens with every object as a default value for a parameter in Python. It's just that with mutable objects, operations performed on them don't result in a new one, so you're left with an edited version of the original object (that's what mutating an object means). Python dictionaries are mutable objects, so the object in each call with the default value is the mutated dictionary throughout the program's runtime (Python assigns the parameter the same object as a default value, so if it is mutated, you get to keep the changes).

[smoky tapir]

how to keep counter for number nodes for doubly linked lists

[willow mason]

can anyone solve this question please

[keen rover]

!dictionary

[limpid smelt]

you just want the answer directly?

no help or suggestions at all?

[sharp junco]

would this sol work as well

import time

arr = list(map(int, input().split(" ")))

st = time.time()
diff = 10**9
for i in range(len(arr)-1):
    curr_diff = abs(arr[i] - arr[i+1])
    if curr_diff < diff:
        diff = curr_diff
    
print(diff)```

(as in being linear time)

[flat hornet]

guys for be developer advanced in python what i need to know?

[haughty mountain]

shouldn't you be able to verify that yourself by looking at the code?

how many times does the inner part of the loop run?

[sharp junco]

the inner part of the loop will run once for each element in the list, except for the last element.

wait what

[haughty mountain]

err

wrong channel for the last thing

[sharp junco]

npnp

[haughty mountain]

so?

[sharp junco]

it is linear time

QED

[haughty mountain]

correct

[sharp junco]

thanks fr

so when testing this and any other code i have, with the max input and time contraints

was just curious how id do that

i have my code already

just wanna see if it satisfies the contraints

which is 10000

and then 9 seconds

have my code already

[haughty mountain]

generate a hard case

[sharp junco]

like u did here?

the command with the max

[haughty mountain]

you know the constraints, you should have some idea of what case would make your code run slow, generate the worst case you can within the constraints

in the previous thing only the array size matters, in other problems the worst case might be harder to know

[dusky field]

wouldnt he need to make a program to calculate the time

and test the constraints?

[haughty mountain]

you can't test the constraints without having some inputs close to the worst case

[keen hearth]

You have two python range objects, xs and ys. Assume they are normalised so that xs.step > 0 and ys.step > 0. How would you go about finding the smallest integer z such that z in xs and z in ys? Also, how would you find out if they intersect at all (that is, set(xs) & set(ys) != {})? I know it has something to do with modular arithmetic 😄

[dusky field]

sorry, you're right

so what would the inputs be

im curious now

[haughty mountain]

I'm not sure, seems like you would want to maximize the number of overall neighbors within 2 steps

[agile sundial]

aren't you just checking the bounds? i don't think modular arithmetic comes in. if they intersect the bounds will overlap (one will start before the other ends)

[haughty mountain]

I think connecting a bunch of nodes with 70 connections each together is the best you can do

but not entirely sure

[haughty mountain]

different steps

[agile sundial]

ah

[haughty mountain]

so gcd ends up mattering

[keen hearth]

E.g. if xs = range(0, 10, 2) and ys = range(1, 10, 2), then they don't share any numbers in common at all 🤔

[haughty mountain]

it's a diophantine equation

[keen hearth]

Oh no

[haughty mountain]

start1 + step1*x = start2 + step2*y

https://en.wikipedia.org/wiki/Diophantine_equation#One_equation

[keen hearth]

Ah right ok 🤔 How would I go about solving for x and y?

[haughty mountain]

allowing for arbitrary x and y, there generally are either 0 or ∞ solutions

(iirc)

what you tend to do is find one solution, and then solve a related equation to get all solutions

specific and homogenous solutions, I think the terms are

but yeah, read the wiki section

[keen hearth]

Looks like the solution is right there. Thanks!

[haughty mountain]

that will give you the family of all solutions, you might need to do some small work to get the one you are interested in

picking the appropriate k

[tender atlas]

what is the best way to calculate operations of graph/path finding algorithms

[willow mason]

i wanted to know what is the most efficient data structure for this

then i can code

in general we can use dictionary

but is there any better way

[languid nymph]

Just to check my understanding of time complexity, this thing is time complexity O(n), right?

[covert thorn]

https://wiki.python.org/moin/TimeComplexity
it should be.
iteration is obviously O(n), but set/dict membership detection can be, in the worst case, O(n) as well (otherwise O(1)). This means the worst-case is O(n^2), with average case of O(n)

the reason for this is that it's technically possible for a hash algorithm to produce the same hash for all of some number of inputs, and that means linear search is required, which is O(n)

although if you found a series of inputs that hashed to the same value, thus making set/dict detection O(n), you could probably submit them as a bug, and the hash algorithm might get changed

[languid nymph]

Thanks

[agile sundial]

it would depend on the elements in array, but probably O(n)

[haughty mountain]

it's trivial to make this happen for ints, because the hash function is basically identity

but for some reason int hashes weren't randomized like str hashes were

[agile sundial]

the reasoning is that it's less common irl to have a lot of ints like you would strings

plus like, the periodicity is like 2^61 or something

[haughty mountain]

2^61 - 1

that's the most trivial way of causing collisions, but there are others

that doesn't need as huge numbers

idk why they didn't just do a sane hash for ints as well

[haughty mountain]

and that's not true for string hashing?

[agile sundial]

i don't think you can apply the idea of periodicity to strings, though. sure if you got 2^61 strings at least 2 would have the same hash

[haughty mountain]

so why was string hashing an issue?

I'm sure it must have been pretty easy to construct collisions

otherwise no-one would have cared

[agile sundial]

they said it was more common than ints, i think

[haughty mountain]

strings are probably more common as a type in general

but that doesn't really justify leaving int with a bad default hash

[agile sundial]

yeah tbh it's a little strange. i think the reasoning is because it's really fast, but like, how much benefit does that give a program?

[haughty mountain]

I don't think modern hash algos have that much overhead

[haughty mountain]

in fact python already has siphash for strings, you could easily use that for ints as well

just hash the bytes that makes up the int digits

[agile sundial]

rust also uses siphash

[urban kiln]

is it worth getting https://cses.fi/book/index.php for learning more abt dsa or is the pdf version enogh?

[sharp junco]

can someone explain this please

i dont understand the step size part of this

from my understanding, if we had string = "Hello World"

then string = [0:6] would output only World, correct?

but then i dont really undertsand the step size digit here

(hw problem)

[urban kiln]

D i think

[sharp junco]

i wanna understand it

[urban kiln]

its empty

[sharp junco]

what

[urban kiln]

bcs without the -1 it would just output 1

char

wait

i think 4

nvm sry

[sharp junco]

uhh

[urban kiln]

i think D bcs without -1 it would output a string but with its doing steps of -1 and idk how to say it

[sharp junco]

wheres the -5 string

[urban kiln]

-5 is just index from end

wait

i cant word it correctly

[sharp junco]

ahhhhhhh

okok makes sense

so -5 in the string xs = "0123456789"
would be 5, since counting backwards from the end of the given string is how we index negative numbers

[urban kiln]

yes

xs[-5] should output 5

[sharp junco]

thanks

i think its none of the above actually 😆

if you're on ur pc try using an IDE to see for yourself

this seems like a trick q ngl

nvm it is D

bruh this is weird

[foggy berry]

im thinking of learning C++ for algos, since its the best language for competitions, what do you guys think? or python is a viable option

[mystic fractal]

python

[granite vortex]

so i appended these random email adresses from .txt file to a list of strings, when i print the entire list, all the adresses have a \n at the end, however when i print them individually this \n doesn't appear, can someone tell me why?

[haughty mountain]

\n is a newline, and you do see them

print prints space separates, but you get a newline then a space

you can use .strip() to remove surrounding whitespace from the string

[granite vortex]

thx

[high sorrel]

from discord.ext import commands

bot = commands.Bot(command_prefix='.')

@bot.command(name='spam', help='Spams the input message for x number of times')
async def spam(ctx, amount:int, *, message):
for i in range(amount): # Do the next thing amount times
await ctx.send(message)

bot.run('TOKEN HERE')

what is wrong in it?

[delicate ore]

Hi guys, I got a problem on my loss function:

def loss(G: nx.Graph, output):
    _loss = 0
    choices = torch.argmax(output)
    for u, v in G.edges:
        if choices[u] and choices[v]:
            _loss += 1
    return _loss

output is from my torch model, and I want to minimize this loss function. unfortunately, var loss is not derivative with output. Is there any way to replace my for loop using torch operations?

[haughty mountain]

#discord-bots

[haughty mountain]

#data-science-and-ml

[fiery cosmos]

should I change this wide format to long ? say add a column with wine,beer,vodka,champagne,brandy

[lucid wyvern]

Hello, stupid me has a question, at my university they want to host a poker bot battle.
I don't think i'm anymore capable enough to participate but the idea still intrigues me

What would be the smartest approach ?
Mine would be to refer look up tables but it seems not efficient at all, a bit dawning

[agile sundial]

lookup tables are the best. you can precompute chances of all the possibilities

[lucid wyvern]

[restive kraken]

Anyone know an Algo to compare the data of two non binary trees. For example compare the following two trees

[woeful crescent]

Hey guys do you have any resources to practices recursion questions

[iron tangle]

Does anyone know of any program where you can find the Big O notation of a function? I have a set I'm working on and would like to check my answers (professor refuses to give answers even though its a practice :/ )

[heavy sinew]

How are you wanting to compare them? Take a value from each node and find the greater one, replacing elements, etc

[glass river]

is bfs [1,2,4,5,9,7,10,3]
and dfs [1,2,5,9,7,10,3,4]?

[lucid wyvern]

Can't you check the elements of the list on the left to see if they are in the list on the right ?

[keen hearth]

Yep. Starting from 1 those are possible BFS and DFS sequences. Obviously it depends on how you go about choosing at each step if you have several nodes to choose from.

For example, if you always choose the successor with the smallest label, then the DFS sequence would be 1, 2, 5, 4, 9, 7, 10, 3.

[iron tangle]

I see. Interesting, thank you

[granite vortex]

is it bad to get there warnings everytime i install a new package?

[hallow river]

Can someone help me do a really simple mvvm javascript code?

I can't find any simple one that can run online

[limpid smelt]

Can you show what command or package you’re getting this with?

[granite vortex]

i was getting tensorflow, but this happens with every package i ever downloaded, i used pip install tensorflow --user

bcs with just pip install tensorflow it wouldn't work

[limpid smelt]

Are you installing in an env or through vscode terminal or through cmd?

[tardy prairie]

negative indexes start from the end, step is the last digit [start: end :step]. indexing starts with 0 (number 0 is at index 0).

negative step goes from left to right ex. 'hello world'[::-1] would return 'dlrow olleh' ('hello world' spelled backwards)

[copper night]

How can I loop through my dictionary to find values > 1?

[agile sundial]

use dict.values()

[granite vortex]

sorry for the late reply i went to sleep, i install through vs code

[harsh hedge]

Is there a way to get a hashing algorithm by a string value?

[dusky trellis]

by "to get" do you mean use or do you mean get a list of the available hashing algos?

[dusky trellis]

If the former, you can use hashlib.new(algo_name)

if the latter, you can use hashlib.algorithms_available

this is documented here => https://docs.python.org/3/library/hashlib.html

[harsh hedge]

Ok thanks this was my question

[limpid smelt]

are you installing it in a virtual env? If not, try installing through cmd

[fiery cosmos]

Guys I have a question

the question isthis - "calculate the length of python string without any inbuilt functions or using for loop or changing the given string's value and without exception handling"

I bet noone can solve this

so simple yet irritating

I solved it

but took me 6hours to do it

[frank bluff]

what about||```py
s = "pineapple"

l = 0
while s[l:]:
l += 1

print(l)

[fiery cosmos]

omg

you're kidding me

[fiery cosmos]

this was my solution
||```py
x=str(input())
i = 0
while True:
if x[-1:i:-1] == x[-1]:
i+=1
break
else:
i+=1
print(i+1)

[haughty mountain]

ctypes isn't builtins

aww, I think I would need id though

!e

import ctypes
len = ctypes.cast(id('this is 26 characters long')+16, ctypes.POINTER(ctypes.c_size_t)).contents
print(len)

[halcyon plank]

@haughty mountain :white_check_mark: Your 3.11 eval job has completed with return code 0.

c_ulong(26)

[snow garden]

guys what does null value do in database?

[fiery cosmos]

how to calculate the proportion of sales ?

in percent

[royal ether]

Hi Everyone,

Question!

If I have a Series which contains something similar to this:
2021-12-12 True
2021-12-19 True
2021-12-26 False
2022-01-02 False
2022-01-09 True
2022-01-16 True
2022-01-23 False

How could I ask Python to extract only the dates of the first occurences of True?

Meaning, my expected result would be:
2021-12-12
2022-01-09

Thanks!

[agile sundial]

divide by total

[mystic dust]

Is there a better way to write this?

def time_taken(self) -> timedelta:
    value = timedelta()
    last_date = None

    for curr_date in self.timestamps:
        if last_date and curr_date.flag == TimestampFlag.stopped:
            value += abs(curr_date.timestamp - last_date.timestamp)
            last_date = None

        if curr_date.flag == TimestampFlag.start:
            last_date = curr_date

        if last_date:
            value += abs(datetime.now() - last_date.timestamp)

        return value

TimestampFlag is an enum and self.timestamps is a list of objects. The object only has a timestamp (datetime) and flag properties.

[surreal nova]

I've been playing with this for a bit, a better way is to improve readability

first I had the logic for calculating the time taken moved into a separate function

def calculate_time_taken(start_date: datetime, end_date: datetime) -> timedelta:
        return abs(end_date - start_date)

function will be triggered everytime TimestampFlag.start is in the self.timestamps list, and the time taken will be added to the value variable.
this would help you avoid the same logic multiple times

here's the full code from my end

from datetime import datetime, timedelta
from enum import Enum


class TimestampFlag(Enum):
    start = 1
    stopped = 2


class Timestamp:
    def __init__(self, timestamp: datetime, flag: TimestampFlag):
        self.timestamp = timestamp
        self.flag = flag


class Timestamps:
    def __init__(self, timestamps: list[Timestamp]):
        self.timestamps = timestamps

    def time_taken(self) -> timedelta:
        value = timedelta()

        def calculate_time_taken(start_date: datetime, end_date: datetime) -> timedelta:
            return abs(end_date - start_date)

        last_start_date = None
        for curr_date in self.timestamps:
            if curr_date.flag == TimestampFlag.start:
                last_start_date = curr_date
            elif curr_date.flag == TimestampFlag.stopped and last_start_date:
                value += calculate_time_taken(
                    last_start_date.timestamp, curr_date.timestamp
                )
                last_start_date = None

        if last_start_date:
            value += calculate_time_taken(last_start_date.timestamp, datetime.now())

        return value


def test_timestamps():
    # Define a list of timestamps with associated flags
    timestamps = [
        Timestamp(datetime(2022, 10, 1, 8, 0, 0), TimestampFlag.start),
        Timestamp(datetime(2022, 10, 1, 8, 30, 0), TimestampFlag.stopped),
        Timestamp(datetime(2022, 10, 1, 9, 0, 0), TimestampFlag.start),
        Timestamp(datetime(2022, 10, 1, 10, 0, 0), TimestampFlag.stopped),
    ]

    # Create a Timestamps object with the defined timestamps
    t = Timestamps(timestamps)

    # Calculate the time taken for the timestamps
    time_taken = t.time_taken()

    # Print the time taken
    print(time_taken)

    # Expected output: 1:30:00


test_timestamps()

[hollow stone]

Hello,

A functional approach could be to use itertools.groupby (https://docs.python.org/3/library/itertools.html#itertools.groupby) on the second field (boolean) and get the first result of each group (and this first field i.e datetime)
Not a difficult implementation/solution, i let you try first ^^

[silver dock]

Hi Guys!

i need help to flatten JSON inside Pandas DF cell and turn it into a new DataFrame.

i tried, among other things, pd.json_normalize(df['col_name'] but i get an error string indices must be integers

when i check cell's type with type(df['col']), i get Series

any ideas on how to unpack the cell's values that look like a JSON and create a new DF?

figures it out

all good

[cursive garden]

I have question Can we really add valiable after object, models.___ , if i get parameter and add after obeject Or should i make if else?

[royal ether]

I ended up using a for loop and copied the initial list twice each time using an incremented shift but I'm curious about the group by approach so I'll try that out!

Thank you!

[sacred rivet]

Heyo! So I've been learning stacks, and part of the applications is conversion of expressions, and this is where I got stuck. The algorithm to convert an infix expression to postfix is quite straightforward. However, infix to prefix is not.

The algorithm taught to us to convert infix to prefix is -

1) Reverse the infix expression, taking care of parantheses
2) Convert the reversed infix expression to postfix
3) Reverse the result of step 2 to get the prefix 

However, when I convert the given infix expression manually, the result is vastly different. For example, for the expression A+B*C/D+E the manual conversion led to ++A/*BCDE whereas the result of the algorithm is +A+*B/CDE

Now I'd like to know where I'm going wrong with this concept. Any help is greatly appreciated.

[hollow stone]

Here a very simple/naive implementation with itertools.groupby tool :

>>> series = [
... ("2021-12-12", True),
... ("2021-12-19", True),
... ("2021-12-26", False),
... ("2022-01-02", False),
... ("2022-01-09", True),
... ("2022-01-16", True),
... ("2022-01-23", False),
... ]
>>> from itertools import groupby
>>> [next(g)[0] for k, g in groupby(series, key=lambda t: t[1]) if k]
['2021-12-12', '2022-01-09']

Enjoy ^^
ps: i'm in love with functional programming and short one-line code 😉

[royal ether]

I will probably have more questions like this!

Thanks for your help!

[wise blade]

    for i in range(1, 3):
        data = createList(i)
        file.write(str(data))

Hello. I am trying to implement a function that writes in a file a list of lists that i create with createList function.
However this function does something like [[1, 2], [3, 4]] [[6, 7], [8, 9]].... And I want something like
[[1, 2], [3, 4], [6, 7], [8, 9].....]
Is there any way to concatenate those newly created lists in the for loop?

[vagrant anchor]

when you don't know when to take a break from typing, your compiler will let you know :v

[half lotus]

Step 1:
Reversing one level at a time, starting from the outer layer and working inwards:

(A + ((B * C) / D)) + E
E + (A + ((B * C) / D))
E + (((B * C) / D) + A)
E + ((D / (B * C)) + A)
E + ((D / (C * B)) + A)

Step 2:
Again, one layer at a time from outer to inner

E + ((D / (C * B)) + A)
E((D / (C * B)) + A)+
E((D / (C * B))A+)+
E((D(C * B)/)A+)+
E((D(CB*)/)A+)+

Step 3:
Should be obvious to see that reversing the result of step 2 gives +(+A(/(*BC)D))E, which is indeed the correct prefix version of the expression.

[fiery grove]

I wanted to ask if anyone knew Solana blockchain development in python

[sacred rivet]

Thanks for the detailed explanation. I think I get where I went wrong. I didn't handle operator precedence in step 1, leading to confusion in step 2 since the precedence is not the same as that in step 1. This helped a lot :)

[half lotus]

No problem. Glad it helped

[lean walrus]

you dont need ( and ) here
++A/*BCDE is also valid

[random kraken]

help

oops wrong

[half lotus]

I know. But it's easier to see precedence with them.

[vital talon]

Is there a built-in Mapping type in Python that can hold arbitrary objects as keys including impossible to hash types? I don't really care if lookups are O(n) and identity can be compared naively with __eq__ but I just want to know if I should make one for myself or one already exists in the stdlib.

[agile sundial]

isn't that just a list, then

if you just have __eq__ (not even __lt__), then you pretty much can only use a list

[vital talon]

Yeah, I would just build it as a list of tuples, I just wanted to know if something that already implements the Mapping protocol already exists. Guess not.

[agile sundial]

yeah or just 2 lists

[ionic crystal]

Hey guys, so I have a doubt...
I've got this huge dataset and it contains 30 parameters which result/affect one target parameter.
I'm supposed to find out the pattern or the logic in which the other parameters affect the target parameter.
Is there any library or package available for it or is there any way I could do it in simple steps? 🙁

[agile sundial]

tensorflow?

[marsh flicker]

im having a preety strange problem

    def updatesensor(self):
        #Variables so it soesn't call the same sensor data more than once
        #Increases speed reduces error
        acc = self.accelerometer.acceleration
        #print(acc)
        ambient_pressure = self.bmp.pressure
        
        self.sensordict = {
            'time': time.time(),
            'acc_x': acc[0],
            'acc_y;': acc[1],
            'acc_y;': acc[2],
            'acc_resultant': (acc[0] ** 2 + acc[1] ** 2 + acc[2] ** 2) ** 0.5,
            'solenoid1': self.solenoid1.value,
            'solenoid2': self.solenoid2.value,
            'solenoid3': self.solenoid3.value,
            'solenoid4': self.solenoid4.value,
            'solenoid5': self.solenoid5.value,
            'gps_alt': self.gps.altitude_m,
            'gps_latitude': self.gps.latitude,
            'gps_longitude': self.gps.longitude,
            'gps_speed': self.gps.speed_knots,
            #'bmp_temp': self.bmp.temperature,
            'bmp_altitude': 44307.7 * (1 - (ambient_pressure / self.bmp.sea_level_pressure) ** 0.190284),
            'bmp_pressure': ambient_pressure
            }

and it returns this

{'time': 1670799768, 'gps_latitude': None, 'gps_longitude': None, 'acc_x': -5.64863, 'bmp_pressure': 1013.99, 'bmp_altitude': 46.8398, 'gps_alt': None, 'solenoid4': False, 'acc_y;': 7.33537, 'acc_resultant': 9.86718, 'solenoid5': False, 'gps_speed': None, 'solenoid1': False, 'solenoid2': False, 'solenoid3': False}

why is it mixing my dictionary up?

[lean walrus]

Because dict are not ordered on old python versions

What version are you using?

Dict are ordered after 3.7 iirc

[lean walrus]

You can use list of 2-tuples for unhashable keys and dict for hashable keys
It will be faster if you have a lot of hashable keys and only several unhashable

[hollow ibex]

Hello, I was wondering if there is an algorithm that given a list of numbers and a target number, the algorithm finds the smallest sum of a list of numbers that is greater than or equal to the target number:

numbers = [2, 4, 5, 3, 1]
target_number = 12

print(find_lowest_sum_numbers(numbers, target_number))

Output:

(2,4,5,1)

Thanks in advance!

[rocky stream]

Hello!
How do you write a for loop in list comprehension where the result depends on an external variable?

cwd = Path()
level = 0
result = []

for row in rows:
  cwd = adjust_cwd(cwd, level, row.level)
  if row.is_row():
    result.append({path: cwd / f"{row.name}.md"})
  else:
    cwd = cwd / row.path
    result.append({path:cwd})
  level = row.level

[frosty plover]

hi guys.

I need some tips before practicing OOP design patterns.
I am a developer for around 5 years - I know OOP in a "general" way, i mean... I don't know what's a "factory" pattern, but will definitely know when and how to use it (i think, and that's why i want to practice).

• Is it recommended to know all patterns names and common uses?
• How is it best to practice these kind of stuff (not a beginner). Just build an "vanilla" python OOP based app?

I would like some kind of a pro review of my app. how do people write code today and make sure that they are doing it "the right way"🤔

[haughty mountain]

#software-architecture

[topaz otter]

py```
def format_name(f_name, l_name):
if f_name == "" or l_name == "":
return "You didn't provide valid inputs."
formated_f_name = f_name.title()
formated_l_name = l_name.title()
return f"Result: {formated_f_name} {formated_l_name}"

#Storing output in a variable
formatted_name = format_name(input("Your first name: "), input("Your last name: "))
print(formatted_name)
why does this work as intended but when I swap it to py
if f_name or l_name == "":

[fiery cosmos]

do SAS plats like AWS utilize parallel processing?

[agile sundial]

it's hard to imagine why they wouldn't

[copper palm]

whats the point of talking about data structures and algoritmes, in one of the slowest languages known to man
whatever you write, an amateur in a language like rust, c++ or golang, can prob make it a few 100 times faster, and prob despite an expontial time complexity
and the other problem is that even if you have a time complexity of say O(n) for an algorithm instead of O(50*n), that doesent mean that O(n) is faster,
if you have terible cache locality or the problem cant be processed in paralel, it could still be allot slower

[agile sundial]

i mean, switching languages is just a constant factor speed increase :P

[copper palm]

so whats your point?

[agile sundial]

the point of learning DSA is not so that you go out and implement them on your own everywhere. your goal is to learn the concepts. in some cases, the language you're writing it in does matter (but not for the reasons you mentioned) in my opinion, but concepts can be learned in any languuage

also, i don't really get your last two sentences, they don't connect with the first part

[copper palm]

whats dsa?

but whats the point of learning these concepts, if your not gonne implement them?

[agile sundial]

so every time you need to sort a list you just write your own mergesort implementation?

the goal is to get exposure to algorithms and learn what tools to use in each situation

[copper palm]

no you just use the builtin sort in python

[agile sundial]

data structures and algorithms

[copper palm]

im pretty sure almost any tool you can imagin already exists in a module in python so why spend the effort to learn all of em, if your not gonne go out of your way to try to implement something faster

[agile sundial]

sure

[copper palm]

you can just black box them, and just learn how to use them, not how they work, that would save time

my terible english

and in the case that you do want to make something faster, your prob just gonne hand that fast algorithm you made, to someone who can implement it as a module instead of doing it yourself

[keen hedge]

hello! I was wondering if anybody had any ideas for using combinations/itertools to get all possible combinations of "event outcomes?" Currently I am putting each outcome into a big list and getting the combinations of those, but it contains tons of invalid combinations with multiple outcomes for the same event.

And filtering them post-hoc is taking way longer than I suspect a native itertools function call would take

[granite vortex]

if working around the math formula for combinations takes too long, look intro libraries such as asyncio and numpy, they improve the performance by a lot

[keen hedge]

getting the combinations themselves is pretty quick its more filtering out the invalid ones thats a problem

but I guess just getting the combination iterator isnt the slow part anyway

its more actually iterating through them all

[haughty mountain]

the point is knowing what kind at algos and data structures are available and how they work

if you don't know they are available you can't use them to solve a problem

if you don't know roughly how they work you won't know why and when something is applicable

And the considerations that goes into picking the right algorithms for a problem.

I see learning about ds&a as adding tools into your toolbox, and familiarizing yourself with how to use those tools

[prime atlas]

Hey! Is there any way to filter a list of lists by lists with a specific column value without using Pandas?

I've looked up on the internet but didn't find anything that satisfies my needs.

Thanks in advance :)

[keen hedge]

I was able to do what I was trying by using Cartesian product rather than combinations. Put the outcomes of each event into a list and then those lists into another list, then get product(*listOfLists)

But it still makes my computer crash. I think there are just too many combinations

[fiery cosmos]

HEX_FILTER = ''.join(
    [(len(repr(chr(i))) == 3) and chr(i) or '.' for i in range(256)])

print(HEX_FILTER)
word = 'Error\t'
map = '..eror'
map2 = 'omk'

print("with hex filter:", word.translate(HEX_FILTER))
print("with map:", word.translate(map))
print("with map2:", word.translate(map2))
# output:
#
# with hex filter:  Error.
# with map: Error 
# with map2: Error

pls help me understand this :/

[acoustic yew]

Hello I'm trying to type an algorithm that basically finds the number of different ways square shapes can fill a random given XY Grid

For example there is only 1 way to fill 1x1 grid and that's by puting a 1x1 square, there are 2 ways to fill a 2x2 grid first is 4 1x1 squares and second is 1 2x2 square, how can i approach this problem without brute force?

[stable pecan]

when you fill a square, do you need to use squares that are all the same size? or does this count as a filling:

##...
##...
@@...
##@##
##@##

(1 3x3, 3 2x2, 4 1x1)

[acoustic yew]

is that a 5x5 grid?

[stable pecan]

yeah

[acoustic yew]

yeah that fits what i need

like that would be a way to fill 5x5

they just have to be squares

[stable pecan]

makes it tougher, but i suppose there's probably some smart way to break it into smaller pieces

[acoustic yew]

i already did find a way

by brute forcing every 4x4

way to fill

which is finding combinations of position each XxY size has to fit

but still working on it

[harsh hedge]

I've written this little code but for some weird reason, when I execute it, the print(Fore.GREEN + "[+] Writing bytes into \"" + file + "\"", end="") gets outprinted after the writing is done

    print(Fore.GREEN + "[+] Writing bytes into \"" + file + "\"", end="")
    
    start_time = get_current_time()
    with open(file, "wb") as f:
        f.write(bytes("1".encode("utf-8") * size))

    time_taken = get_current_time() - start_time
    print(" Done(" + format_time(time_taken) + ")")

[river inlet]

Have you tried an f-string?

[harsh hedge]

How do you mean this?

[river inlet]

https://datagy.io/python-f-strings/

[harsh hedge]

This didn't fix my problem

[harsh hedge]

Ok I found out that this is because of I print it with end=""

But why? I need to stay in the same line
What should I do

[austere flame]

I did ANOVA two-way hypothesis testing and made this mistake : ValueWarning: covariance of constraints does not have full rank. The number of constraints is 9, but rank is 1

I have no null value in my data, but I don't know how to fix it

[woeful cove]

@codeine#1240

[limpid smelt]

did you find the solution?

[harsh hedge]

Yes. You have to set flush=True, because normally the buffer is only flushed after a new line or longer string

[foggy lynx]

Hi guys--I have an operations research question

Is there any Python library, that given a convex set of constraints, can just generate N random feasible solutions?

I.E. I have a non-convex objective, and do not wish to use mixed integer programming due to computational costs in general, so instead, would like to simply enumerate 1000 random feasible solutions, and take, say, the top 20 or so in order to feed into a convex optimization problem.

Is there some exiting Python library that'd allow me to do this?

[quaint halo]

Pyomo is pretty good for optimization problems

[nimble coral]

What do u guys think about 4/6/8 month-1 year data science bootcamp? Especially for undergraduate IT student

[old rover]

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        l, r = 0, 1 #left is buying, right is selling
        maxP = 0

        while r < len(prices):
            if prices[l] < prices[r]:
                profit = prices[r] - prices[l]
                maxP = max(maxP, profit)
            else:
                l = r
            r += 1
        return maxP

i don't understand the point of l = r

as opposed to l += 1

https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        hashset = set()

        for n in nums:
            if n in hashset:
                return True
            hashset.add(n)
        return False

what's the advantage of using a set instead of a list?

and also, don't sets contain only unique items?

https://leetcode.com/problems/contains-duplicate/

actually it works because sets only store unique items

i think i figured it out

[agile sundial]

well this solution doesn't depend on sets only having unique items. it uses a set because it's fast at checking if it contains an element or not

[old rover]

so then

if i used a list it would still work?

[agile sundial]

yea

[old rover]

weird, bc a list doesn’t work for the test cases

[agile sundial]

why

probably speed issues

[old rover]

yes

it says it timed out

[agile sundial]

but it is correct

[barren skiff]

Wait

Lists times out?

That is strange

N only goes up to 1e5 and the solution is O(nlogn)

[agile sundial]

it's n^2

[inland tusk]

I have a begineers problem here with python and pandas. Trying to get the example running. Heres all I've done in the dir.

mkdir pandas; cd pandas;
python -m venv .
bin/activate
pip3 install -r requirements.txt
python3 pandas.py

This is where I get the error

ModuleNotFoundError: No module named 'pandas.core'; 'pandas' is not a package

My requirements file is pretty simple

yfinance
numpy
pandas_ta

Any idea what I'm doing wrong here?

[harsh hedge]

I want to dynamically split up a task into n threads. How can I get n depending on how much the hardware of the pc can take

[vocal gorge]

A bit of googling of python cpu count later, looks like there's a builtin for that, https://docs.python.org/3/library/multiprocessing.html#multiprocessing.cpu_count

[agile sundial]

also i think ThreadPoolExecutor uses that as the default

[vocal gorge]

processpoolexecutor, presumably

[agile sundial]

yeah something like that

[barren skiff]

Wait am i missing something

The sol with lists sorts the lists

Which is O(nlogn)

[agile sundial]

not his solution with lists

[barren skiff]

And the linear checks the list for duplicates

Oh

I thought u were saying that lists are too slow for this problem

My mistake

I misunderstood

[fiery cosmos]

hello guys i want to learn data structure via python, i want complete topics like linked list, trees, dijkstra algo and everything please suggest me any course or youtube channel

[harsh hedge]

Im making a script with which you can create files with a given size, I add the bytes together with size * '000000000000000000\n but this takes way to long. What can I do about it?

[lean dome]

Do the contents need to be ASCII '0' characters, or is anything that takes up space good enough? If you just need to take up space, open the file in binary mode ("wb"), seek to the desired size with seek, and then close the file

[austere sparrow]

Does it have to be a python script?

You can use fallocate (at least on unix)
https://unix.stackexchange.com/a/381339

[urban kiln]

do trees appear often in contests(not leetcode)

im thinking about skipping trees for now since i havent seen any tree problem except in leetcode contests

[fallow garnet]

How can I make a hash function that's around as fast as the inbuilt hash function?

Also without using any other libraries

[calm verge]

You might not see trees directly but for many problems on there it's helpful to think of it like a tree and do recursive traversals on it

[calm verge]

I'd assume the inbuilt one is implemented in C for all the primitive types, so if you wish to do so in Python that would always have extra overhead

[fallow garnet]

I see, so what would be the best hash function even if it's not as fast as the inbuilt one?

[calm verge]

This is hard to answer generally tbh. What's your use case? You can have a simple but fast hash function which might work well for certain types of data but not others

If you want a good general purpose hash function there's a few common ones you can find, I don't know about their performance in python

[fiery cosmos]

okey so i am reading https://runestone.academy/ns/books/published/pythonds3/Introduction/ObjectOrientedProgramminginPythonDefiningClasses.html tthis and i have trouble understanding concept of Connector implementation

so basically this ```
g1 = AndGate("G1")
g2 = AndGate("G2")
g3 = OrGate("G3")
g4 = NotGate("G4")
c1 = Connector(g1, g3)
c2 = Connector(g2, g3)
c3 = Connector(g3, g4)
print(g4.get_output())

Enter pin A input for gate G1: 1
Enter pin B input for gate G1: 1
Enter pin A input for gate G2: 1
Enter pin B input for gate G2: 1
0
``` this is output

how is there only 2 and gates involved?

[fallow garnet]

I need a hash function for a hashset implementation and it only really needs to do strings

[fiery cosmos]

nvm got it

[calm verge]

Even for strings it depends. Here's some you can look at:
XOR hashing
DJB2 hash
Murmur Hash
Superfast hash
Jenkins hash

Try them out, see which one works faster / better. They're all mostly straightforward to implement

Is there a reason you're not just using the built-in hash function though?

[fallow garnet]

What do you think of this code?

def fast_hash(string):
    h = 0
    for c in string:
        h = (h * 31 + ord(c)) & 0xffffffff
    return h

[fallow garnet]

Cuz I'm required to build my own for the project

[calm verge]

Might work 🤷‍♀️ again you gotta remember how good a hash function is depends very much on the input

I don't know what sort of strings you have. You should just implement a couple and see how many collisions you're getting with each

[fallow garnet]

Also one thing to note is that this has way better distribution compared to the inbuilt function?

How is that possible

[calm verge]

For your particular input, perhaps

[fallow garnet]

I see

[calm verge]

It's hard to talk about absolutes without doing some mathematical analysis on what the distribution would actually be like

[fallow garnet]

Well I'm gonna be using a really huge txt file with words to test it

Like around 15 million words

Just normal English words

[calm verge]

Just try it out, if it works it works

If there's something more specific you're looking for we can talk, but I can't tell you what is "better" without context

[fallow garnet]

I see

I cannot run it on the main test file

But I got a smaller sample

Which I can test using my phone

First run uses the inbuilt while the second uses the code I showed

[calm verge]

2000 is not a lot of words, you probably want to do more precise timing with something like timeit module to see actual performance comparison

[fallow garnet]

The file has 13k words, it just shows the amount of unique words

[calm verge]

Ah I see

[fallow garnet]

I've never used the time it module so I'll try it

[tacit thorn]

can someone explain this algorithm to me

def encrypt(plaintext, key):
alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
ciphertext = ""
for i in range(0,len(plaintext)):
print('1.',ciphertext)
character = plaintext[i]
print('i = ', i)
ciphertext = ciphertext + character
for j in range (0,key):
ciphertext = ciphertext + random.choice(alphabet)
print('2.',ciphertext)
return ciphertext

i dont understand how to decrypt it

[urban kiln]

oh ok i guee ill practice them more

[urban kiln]

why is inserting after a Node in a linked list O(1) and not O(n)?

dont u have to get to the given node before you can make given_node.next = ...

[agile sundial]

inserting after a node is O(1). inserting into a linked list when you don't have a pointer to a node is O(n)

[urban kiln]

ah ok

is this important to know how to handle?

"multilevel doubly linked list"

[agile sundial]

i've never seen that before

[urban kiln]

i hope ill never see it again

it looks terrible

"""
# Definition for a Node.
class Node:
    def __init__(self, val, prev, next, child):
        self.val = val
        self.prev = prev
        self.next = next
        self.child = child
"""

[calm verge]

Sounds like a binary tree sideways to me

[agile sundial]

it's not a tree it's directed

[urban kiln]

yea

[urban kiln]

idk

ill try to solve it later

ez

[urban kiln]

oh i didnt know that there is a penalty

oh

[urban kiln]

im trying the second rn

i was able to implement a way to solve some cases

but now i have to figure out how to solve the rest

omfg i cant solve ot

and time is running

[urban kiln]

damn i was so close to solving it

[fiery cosmos]

That for https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/? It's a neat challenge but not really a data structure to worry about in the real world

[urban kiln]

ok

[mint zephyr]

any youtube playlist recommendations for into to dsa with python?

[subtle ravine]

anyone wanna help me debug some pytorch code

[copper hawk]

I think this would more appropriately fit #data-science-and-ml

[fiery cosmos]

Can anyone here help me with a discrete problem really stuck on it?

[fiery cosmos]

hey, short question due to lack of my knowledge (started coding yesterday)... Trying to implement an automatic option to open an app on mac... but i have no clue how to input the app correctly...
Currently stuck on: subprocess.popen (/application/xxx.app)... Not a single clue how to do it correctly...

[river shell]

Json some expert here

?

[austere flame]

Hi everyone, does any one know any function that print each permutation of a list without saving all of them

I find some functions but have problems with memory

[vocal gorge]

itertools.permutations?

[quaint perch]

what is the formula for load factor in hash table

[quaint perch]

oh nvm ik it

can someone review my hash table implementation

[halcyon plank]

Hey @quaint perch!

It looks like you tried to attach a Python file - please use a code-pasting service such as https://paste.pythondiscord.com

[quaint perch]

https://paste.pythondiscord.com/maroyavapo

[austere flame]

thanks, i will try

[ivory fern]

help me please

[dapper sapphire]

I'm interested in learning to optimize sql.

Do leetcode sql questions deal with sql optimization?

[rare prism]

Anyone know of any more efficient versions of the diffusion library?

There are a few issues like it not being very good at smartly allocating VRAM

[languid nymph]

    mid=(len(a))//2 #+1
    start=0
    while start!=mid:
        if a[start]>a[mid]:
            a.insert(start,a[mid])
            a.pop(mid+1)
            mid+=1
        start+=1
    return a```
Time complexity of this (dependent of number of items in list a would be O(n), right?

[lean walrus]

No, it is O(N^2)

[languid nymph]

I was just browsing through some website with explanation of big O notation and I think it's actually O(n). Mid variable starts at the the middle of array a. Then in a/2 iterations, it can get up to len(a). a/2 iterations will also boost start to a/2. Then only a/2 iterations left, for start to get to mid, where the loop ends. Sorry, I've probably explained the script wrong. I will try to run few checks to confirm the O(n)

It seems to have linear time complexity. Sorry for the confusion. There were also few bugs so the code I sent was wrong

[runic tinsel]

insert is not constant time

[lean walrus]

pop is also not

[languid nymph]

Oh :/ Thanks for letting me know

[old horizon]

Anyone here familiar with numba prange that can help?
I'm wondering in variables defined inside a prange are independent between loops.

https://discordapp.com/channels/267624335836053506/1055466095085109248/1055466095085109248

[worldly jolt]

im having some trouble with a bit of code,
assuming I have 6 points how can I find 2 triangles such that the sum of their area is maximized?
I need this to be moderately fast as I plan to loop over this n^2 times so going over all combinations/ permutations is to slow do to repeated triangles like ((0,0)(0,1)(1,0)) and ((0,0)(1,0)(0,1)) that use the same points in different orders

[sand iron]

You should just loop over the combinations not all the permutations
so ⁶C₃ is 20 which should be small enough?

then take the largest two elements. You could either use sorted or heapq.nlargest if you wanna be fancy, not sure what would be faster at such small scales

[untold torrent]

Can anyone explain what Im doing wrong in this post order traversal code

[gaunt forge]

Practice, practice, practice. There's really no substitute.

[worldly jolt]

it was bothering me that is was going over ther same triangles twice or more, so I ended up "pining" the first point to triangle_A and looping over combinations of length 2 in the remaining 5 points, thanks for the help 🙂

[vital scaffold]

class Queue:
    def __init__(self):
        self._list = [None] * 10
        self._size = 0
        self._front = 0


    def enqueue(self, element):
        if self._size == len(self._list):
            self._resize(2 * len(self._list))

        end = (self._front + self._size) % len(self._list)
        self._list[end] = element
        self._size += 1


    def dequeue(self):
        answer = self._list[self._front]
        self._list[self._front] = None
        self._front = (self._front + 1) % len(self._list)
        self._size -= 1

        if 0 < self._size < len(self._list)//4:
            self._resize(len(self._list)//2)
        return answer


    def print(self):
        for i in range(self._front, self._size + 1):
            print(self._list[i])


    def _empty(self):
        return self._size


    def _resize(self, size):
        new = self._list
        self._list = [None] * size
        walk = 0
        for i in range(len(new)):
            self._list[i] = new[walk]
            walk = (walk + 1) % len(new)
        self._front = 0

qq = Queue()
qq.enqueue(5)
qq.enqueue(5)
qq.enqueue(5)
qq.enqueue(5)
qq.enqueue(5)
qq.enqueue(5)
qq.print()

qq.dequeue()
qq.print()

The dequeue() function is returning None But it should return 5.Why it is that ?

5 5 5 5 5 5
None
5 5 5 5 5 

[sand iron]

Not sure what you mean, iterating over the combinations will ensure you don't go over the same triangle twice.

[worldly jolt]

yes but I have 6 points, so for each 3 I iterate over I also need the other 3 points to create a triangle
ie - 1,2,3,4,5,6
iterate over 1,2,3
second triangle will be 4,5,6

and then when I iterate over 4,5,6 the second triangle is 1,2,3

and the order is irrelevant so I avoided these computations

[sand iron]

If you hold the state you'll avoid having to iterate over triangles you've already seen

[worldly jolt]

what do you mean by that?

[main kiln]

does anyone know altair python?

[sand iron]

Let's say that the triangles are stored in a list[tuple[int,int,int]] and let's call it points and let's say you have an area function that has a signature (tuple[tuple[int, int], tuple[int, int], tuple[int, int]]) -> int.

Now I am saying that the following code will only go through 20 triangles without revisiting ones already seen.
max_area = sum(heapq.nlargest(2, map(area, combinations(points, 3)))

[worldly jolt]

I am not looking for the 2 largest triangles, im looking for the pair with largest sum

[sand iron]

The pair with the largest sum will be the two largest triangles from all possible triangles.

[worldly jolt]

*matching pair

from 1,2,3,4,5,6
I pick 1,2,3
then 4,5,6 are not picked
sum(area(1,2,3),area(4,5,6)) is what im trying to maximize

1,2,3
1,4,5
are not valid as 1 appeared in both triangles

or because 6 is not used

depending how you look at it

[sand iron]

You didn't mention this initially

[worldly jolt]

yea, this was unclear sry

my current code looks like this

best = []
size = -1
for i in itertools.combinations(range(5), 2):
  trigA = [points[i[0]], points[i[1]],points[5]]
  trigB = [points[ii] for ii in range(5) if not ii in i]
  if (tmp := triangle_list_area([trigA,trigB]) > size):
     best = [trigA,trigB]
     size = tmp
return best

where points is a list of 6 Tuples(int,int)

[sand iron]

You've precomputed the areas?

[worldly jolt]

sry im having trouble copy pasting as the code is indented

triangle_list_area is computing the area itself

[sand iron]

triangle_list_area computes the area of two triangles?

[worldly jolt]

any amount but for our case yes

[sand iron]

try using tuples where you know the data is immutable

points_id = set(range(6))
max(map(lambda c: (triangle_area(*c), triangle_area(*(points_id - set(c)))), combinations(range(6), 3)))

try something like this

[worldly jolt]

I thought about using sets, but if I have the same points twice this method breaks

and a dictionary is getting a bit ugly

although I might come back to that as I need to make a version for 9 points

[sand iron]

but if I have the same points twice this method breaks

In that case try using distinct_combinations from the more_itertools lib.
https://more-itertools.readthedocs.io/en/stable/api.html#more_itertools.distinct_combinations

[worldly jolt]

this would actually also break

[sand iron]

When you use distinct_combinations you'll iterate over points rather than indicies.

[worldly jolt]

it just means all triangles with a duplicate in them would count as 0, but there are some options where I split the duplicate across the two triangles

I always iterate over points as calculating indicies is slower

[sand iron]

You're bringing in new constrains as we speak.
I think at this point see what works best for you 👍

[worldly jolt]

not a constraint, just what I did in my code

if it can be done with indicies it's also good

for example the points (0,0,0,0,1,2) would still have a valid option in 0,1,2 0,0,0

[sand iron]

well first up the points will be tuple[int, int] not int
and that's exactly what distinct_combinations would do

anywho, as I said before do what works best for you

[worldly jolt]

I see, I thought it was tossing out similar values where it is tossing out permutations that contain the same variables, thenks 👍

[heady obsidian]

ayo can anyone suggest a website that teaches dsa courses for free

kinda urgent...

[plain folio]

Can anyone explain to me the difference between LOESS and MARS algorithms? From what I understand they both split up datasets and run regression techniques on the subsets of data to get a more accurate measure.

[haughty mountain]

why not normalize the triangle points, e.g. sort them

[worldly jolt]

how would that help tho?

[fiery cosmos]

bunu nasıl çözerim

[worldly jolt]

I nums is a list of strings, you need to change it into a list of ints

nums = [int(i) for i in nums]

or

for i in range(len(nums)):
  nums[i] = int(nums[i])

first one is better, but second one uses more basic syntax

[fiery cosmos]

where should i add this

[steady finch]

send me the code

ill fix it

[fiery cosmos]

@steady finch thankss

bro

[steady finch]

you're welcome

[urban kiln]

nums = list(map(int, input().split()))

[urban kiln]

i think just mapping the input would be even better

[worldly jolt]

I know of the other ways but considure your target audiance

[urban kiln]

oh ok

[worldly jolt]

also this is nicer I think:

[int(i) for i in input().split()]```

[urban kiln]

imo not but ig its just preference

i just use map bcs it can be used without lists too

for example when knowing that u will get 3 ints as input, u could do:
int1, int2, int3 = map(int, input().split())

nvm

that could be done with list too

ig there is no big advantage

but typing () is faster than []

oh but you could just do tuple too

ig there are no advantages

nums = list(map(int, input().split())) -> 38 chars
nums = [int(i) for i in input().split()] -> 40 chars

nums = sorted(map(int, input().split())) -> 40 chars
nums = sorted([int(i) for i in input().split()]) -> 48 chars

[worldly jolt]

a,b,c = [int(i) for i in input().split()]
would also work
the reason I prefer the comprehension is that it is one tool to do the work of both map and filter as well as the existence of one line generators, set comprehensions and dictionary comprehensions
people prefer it as it is one tool to do all of these rather then like 10 tools

[urban kiln]

ik

[urban kiln]

https://leetcode.com/problems/reward-top-k-students/

can someone with leetcode premium send me a writeup of this question

cant check what i did wrong since it is a premium question

[heady obsidian]

im prefer a website

[urban kiln]

you cant go to the discussion/solution section of the sectoon

atleast that was the case when i asked

and it said that i needed premium to access the question

(i tried viewing the solution page 3minutes after biweekly ended)

[urban kiln]

but ty

why are you converting the pos and neg list to sets

i thought the list is already not containing duplicates

og

oh

my bad

[bronze gorge]

Hello guys I need a good resource for DSA?

I'm a beginner.

[urban kiln]

how can i measure an angle, knowing the coords of points which are connected to each other

[opal oriole]

https://en.wikipedia.org/wiki/Law_of_cosines

[urban kiln]

part of the problem 💀

probably the easiest sub problem

[fiery cosmos]

???

i read but i don't understand

[urban kiln]

i think you forgot an indentation block

just put a space before the red marked line

[rare moss]

this is python pseudocode, i wanna ask what property of OOP is in the Pink shape image?

[cinder cave]

They are just calling Class methods

[earnest cave]

this is a leetcode question , in the example 1 , why is the answer not 10 , as we can start from index 0 , and pay 10 to climb 2 steps as allowed , and reach third , so according to me the answer should be 10.

I am not able to understand the problem statement , i guess , completely?

please help

[covert thorn]

the top of the stairs is not the last element of the list, but right after. i'm not sure that i agree with that interpretation of the definition of steps, though, and they probably should be more clear

[earnest cave]

ok , i am also not comfortable with interpretation , thanks for help

[rare moss]

how do I even test this

and idk what X[100] means

[covert thorn]

100-element array/list of integer type

[rare moss]

how do you test the output

[covert thorn]

easiest way is to port the pseudocode into a language of your choice (python, for example) and run it

other way is to recognize what the algorithm is doing and do it all in your head

[rare moss]

also how to port X[100] to python

I know some conversion like

N = int
K = int

N = 4
K = 1
while K <= N:

[covert thorn]

probably X = [0 for _ in range(100)]

[rare moss]

but idk about X[100] or something like array

what is the '_' ?

[covert thorn]

the first two line of the pseudocode, declaring variable types, are irrelevant to python. ignore them.

[covert thorn]

it's a symbol commonly used for when you don't care about a variable's actual value

[rare moss]

what about X[N/2], will it becomes X = [0 for _ in range(N/2)]
?

[covert thorn]

no, that's accessing the value within X with the index N/2. although you would have to ensure N/2 stays an integer, so you would need X[int(N/2)] or X[N//2]

!e

l = ['a', 'b', 'c', 'd']
print(l[3])

[halcyon plank]

@covert thorn :white_check_mark: Your 3.11 eval job has completed with return code 0.

d

[rare moss]

k imma try to convert them all

!e

N = int
K = int
X = int
X = [0 for _ in range(100)]
N = 4
K = 1
while K <= N:
    X[int(K)] = K**2
    K = K + 1
print(X[int(N/2)])
print(X[1], " ", X[int(N-1)])

[halcyon plank]

@rare moss :white_check_mark: Your 3.11 eval job has completed with return code 0.

001 | 4
002 | 1   9

[rare moss]

is this correct?

[covert thorn]

almost:
it's not clear why the intended behavior of Write is, but judging by the single answer line, i'm assuming it's referring to a language with no implicit line breaks on print calls. It's also not clear if the second Write call contains a space in the middle or not (" " or ""). With the space would probably be 41 9 and without 419.
If it does use implicit newlines, then

4
1 9
``` or

4
19

tl;dr maybe `41 9`?

it's kinda of a terrible question, tbh

[rare moss]

how is it 41 9

print(X[int(N/2)])should be in the first line

I mean how 4 and 1 be concatenated

[covert thorn]

many languages do not automatically add a newline in their print statement. python does, but, for example, C/C+++ and Java don't, so ```c
printf("a");
printf("b");

would result in `ab` instead of

a
b

The problem is that it's not clear what `Write`'s expected behavior is

[rare moss]

write is print in python

correct?

[covert thorn]

both write output to screen, yes. Exactly how they do that may differ

This is a, what, homework assignment? I would suggest asking the professor or such how Write is expected to behave with regard to newlines

[rare moss]

I got it from course, basic pseudocode

I try to convert them all to python

[fiery cosmos]

Prepare the program that will do the following for the y value to be entered from the keyboard. 1²+2²+3²+4²+……. +y²

how can I do it

[old rover]

def isAnagram(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False
        countS, countT = {}, {}

        for i in range(len(s)):
            countS[s[i]] = 1 + countS.get(s[i], 0)
            countT[t[i]] = 1 + countT.get(t[i], 0)
        for c in countS:
            if countS[c] != countT.get(c,0):
                return False
        
        return True

i don't understand countS.get(s[i],0)

.get gives the value for a key

0 is the default value

i've just never seen it written like that

also, why can't we write if countS[c] != countT[c]?

neetcode says it's to avoid a key error

because if the key doesn't exist in countT, then it would give a key error

[tepid pivot]

Am looking for a one teammate or two by max so we can learn together and solve ex. From leetcode everyday by choose a specific hour. I just start to attend udacity course for algorithms and data. Anyone interested DM me

[tepid pivot]

Ok add me, yes i need just a few people so we can have a voice chat while we are solving the exercises

[round valley]

def is_anagram(a, b):
    return sorted(a) == sorted(b)

[fiery cosmos]

any python developers here? need help! i am using nsetools library to fetch top gainers and losers. i wanto fetch the data for every 5 minutes. how can i do that?

[old rover]

that works too

[old rover]

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        res = defaultdict(list) #mapping character count of each string to the list of anagrams

        for s in strs:
            count = [0] * 26 #a ... z

            for c in s:
                count[ord(c) - ord("a")] += 1
            
            res[tuple(count)].append(s)

        return res.values()

https://leetcode.com/problems/group-anagrams/

so count is a list of 26 numbers?

what's the point of using ascii values?

[old rover]

i'm confused

but i'll look at it again tomorrow

[haughty mountain]

it's a bit iffy to call it O(1)

I would add in another parameter, alphabet size

otherwise you can do...very silly stuff

[royal torrent]

Is this the channel to ask about leetcode?

I see there is no leetcode channel here
leetcode doesn't even have a discord server

[haughty mountain]

if we changed the alphabet to be unicode or even bigger the thing would technically still be constant, but it's a bit misleading since that constant is now huge

so whenever an alphabet size is involved I much prefer adding it as a parameter

e.g. there are a bunch of algos that are O(n+k) or O(n*k) where n is size of input and k is alphabet size

yes, my point is you're missing a scaling that exists in the problem, it scales with alphabet size

your n will easily fit in an 64 bit int, so the input is O(1), clearly

[teal basalt]

I saw it

Should delete the msg

[fiery cosmos]

also there's such thing as +=, instead of x = x + y, x += y

[teal basalt]

kk just doing pascal work in python to see the answer

lol

[haughty mountain]

I had an argument that by extension said that "there are only a constant number of floating point numbers", so O(1)

[royal torrent]

how come

[haughty mountain]

my point is just that if you don't include the relevant parameters in your complexity it's kinda garbage

[royal torrent]

if i ever feel the need to i would
if channel not dedicated for it then no point

[haughty mountain]

e.g. how fast is looking up a string in a hash table?

no, O(length of string) is correct, and I see people all too often completely gloss that fact over

[haughty mountain]

i.e. to make the lookup you need to hash the object, which is linear in the size of the object

I mean, I do have the role

so yes. I finished on the 25th

algorithmically easier, more impl focused a lot of the time

let's say codeforces, but really any competitive programming thing

advent of code is just a different style of task, a lot of the time the algos involved aren't too interesting

(though a handful of problems per year ends up being interesting on that front as well)

it would be quite boring to me if I didn't spice it up somehow, so I don't do it in python or C++ that I'm really used to

[old rover]

hahaha after looking at two sum for four days i think i finally understand the solution

and it was all about figuring out what my dictionary had, because the dictionary had integers mapped to indices

[old rover]

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
       res = defaultdict(list)

       for s in strs:
           count = [0] * 26

           for c in s:
               count[ord(c) - ord("a")] += 1
            
            res[tuple(count)].append(s)
        
        return res.values()

IndentationError: unindent does not match any outer indentation level
                               ^
    res[tuple(count)].append(s)
Line 11  (Solution.py)

is it because the res[tuple(count)].append(s) isn't perfectly in line with the for?

yes, and bc the return isn't completely inline with the default dict

leetcode was being weird there

i don't understand what the keys of my dictionary would look like tho

the key would be a list of 26 ints

i'm confused

count[ord(c) - ord("a")] += 1
``` i think that's what i don't understand

like how does the string "eat" end up [1, 1, 0, 0, ..., 0] ?

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
       res = defaultdict(list)

       for s in strs: #iterates through strings
           count = [0] * 26

           for c in s: #characters of strings
               count[ord(c) - ord("a")] += 1
            
           res[tuple(count)].append(s)

       print(res.keys())
    
       return res.values()

i put a print(res.keys()) here

dict_keys([(1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0), (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0), (1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0)])

got this output

["eat","tea","tan","ate","nat","bat"]

leetcode used that input

i notice how the keys are different tuples

but it doesn't help me figure out how it's being calculated

[old rover]

how the keys of the dictionary are being formed

sure, but how does that build it?

it calculates the ascii values and then what

[keen karma]

can someone help

what is the answer to this

[urban kiln]

what type of knowledge is usually required for heuristic(i think thats how its spelled) contests

i feel like the problems are completely different from ABCs or codeforces contests

[hollow oasis]

So I have an numpy Poly1D which returns complex numbers. The real part represents X and the Imaginary part represents Y. How do I sample points on the resulting graph?

[old rover]

i'm doing the blind 75

oh i see what you're saying

but after two sum comes group anagrams anyways so

well i need to practice anyways

[hollow oasis]

?

[old rover]

at least i knew to use a dictionary for group anagrams

hey, i think i'm able to understand group anagrams now

i took the test case of "eat"

for example, the letter "e" is the fifth letter of the alphabet

so that's why the fifth position in the list is 1

the letter "a" is the first letter of the alphabet, so that's why the first position in the list is also 1

and t is 20th, so the 20th position in that list is 1

[maiden cradle]

so i have a small problem with json files, if i can ask here idk really, i want to get a data called with a name in the json file and just load it into my python file, so i can use it or just for example print it,
heres the json file :

{
    "SwitchKey": "p",  // this is what i want to get from the json file
    "ExitKey": "y",
    "ReloadKey": "i"
}

so how do i get that into my python file and print it
from what i gathered it has to be something from json.load()

[tepid needle]

!d json.load

[halcyon plank]

json.load

json.load(fp, *, cls=None, object_hook=None, parse_float=None, parse_int=None, parse_constant=None, object_pairs_hook=None, **kw)```
Deserialize *fp* (a `.read()`-supporting [text file](https://docs.python.org/3/glossary.html#term-text-file) or [binary file](https://docs.python.org/3/glossary.html#term-binary-file) containing a JSON document) to a Python object using this [conversion table](https://docs.python.org/3/library/json.html#json-to-py-table).

*object\_hook* is an optional function that will be called with the result of any object literal decoded (a [`dict`](https://docs.python.org/3/library/stdtypes.html#dict "dict")). The return value of *object\_hook* will be used instead of the [`dict`](https://docs.python.org/3/library/stdtypes.html#dict "dict"). This feature can be used to implement custom decoders (e.g. [JSON-RPC](https://www.jsonrpc.org) class hinting).

*object\_pairs\_hook* is an optional function that will be called with the result of any object literal decoded with an ordered list of pairs. The return value of *object\_pairs\_hook* will be used instead of the [`dict`](https://docs.python.org/3/library/stdtypes.html#dict "dict"). This feature can be used to implement custom decoders. If *object\_hook* is also defined, the *object\_pairs\_hook* takes priority.

[tepid needle]

or json.loads

[maiden cradle]

i still didnt get how to get my data out of that

am i looking at it not right

cuz i am not new to python i have been working with python for 3 years

[tepid needle]

d = json.loads(...)
print(d["SwitchKey"])

[maiden cradle]

OHHHH

im so stupid

i thought you do like json.load("SwitchKey")

its 2 AM im tired sorry

[tepid needle]

!e ```py
import json
data = """
{
"a": "b",
"c": "d"
}
"""
parsed = json.loads(data)
print(parsed["c"])

[halcyon plank]

@tepid needle :white_check_mark: Your 3.11 eval job has completed with return code 0.

d

[maiden cradle]

oh 3.11 right now sucks for me, i cant debug

i tried everything to fix it btw

[tepid needle]

!d json

[halcyon plank]

json

Source code: Lib/json/__init__.py

JSON (JavaScript Object Notation), specified by RFC 7159 (which obsoletes RFC 4627) and by ECMA-404, is a lightweight data interchange format inspired by JavaScript object literal syntax (although it is not a strict subset of JavaScript 1 ).

Warning

Be cautious when parsing JSON data from untrusted sources. A malicious JSON string may cause the decoder to consume considerable CPU and memory resources. Limiting the size of data to be parsed is recommended.

json exposes an API familiar to users of the standard library marshal and pickle modules.

Encoding basic Python object hierarchies:

[maiden cradle]

i got it, thanks lad

[maiden cradle]

for now, fair winds my friend

[tepid needle]

since version 2.6

[maiden cradle]

what

wha

[urban kiln]

167. Two Sum II - Input Array Is Sorted

can i just do the same hashmap approach like in part 1?

[haughty mountain]

they want you to only use constant extra space

[urban kiln]

oh hmm

is 2pointer constant space?

[haughty mountain]

it is

[urban kiln]

n log(n) is faster than n^2 right?

yea it is

ig binary with 2pointer

[grim tendon]

is this a good channel to talk automation? or which channel is best

[hardy imp]

I'm not 100% sure if this is the right channel to talk about big O notation but i can't see a better one

if I know the time complexity of a function and i know how long it takes for a certain input n

can i work out how long it will take for a bigger input

for example if an O(2^n) function takes 2 seconds to run at n=2, does that mean it will take 512 seconds to run at n=10?

or is that not how that works

[urban kiln]

2pointer exceeded time limit so i just did hashmap approach:

class Solution:
    def numPairsDivisibleBy60(self, time: List[int]) -> int:
        pairs = 0
        time = [i%60 for i in time]
        hashy = {}

        for i in time:
            if i == 0: 
                if 0 not in hashy: hashy[0] = 0
                pairs += hashy[0]
                hashy[0] += 1
            else:
                if 60-i not in hashy: hashy[60-i] = 0
                if i not in hashy: hashy[i] = 0
                pairs += hashy[i]
                hashy[60-i] += 1

        return pairs

it can be improved i think

i did i==0 since 60-0 is an edge case

and i did the list comprehension to not do modulo in every loop

oh

i used hashmap

all the times%60

i thought thats what i was supposed to do

isnt that the same

i mean the only difference is that you do modulo in every loop

ur solution is way shorter

oh

ok

ty

[urban kiln]

ill try

but idk how to solve sliding window problems(this ones seems to be one)

ill have to look up some concepts for that

ill finish this https://leetcode.com/problems/zigzag-conversion/ first

no, i havent finished school yet(idk if my school would be called high school in us)

you?

i wanted to do it but i havent started yetz

is it difficult?

isnt numRows==len(s) also just return s

[ruby grove]

hi 👋 guys, i want to build a search for my website, it has many articles, every article has json file that includes {url of the page, title, article, keywords, date_created, tags …..} how can i make a text search that can get me the article which is most similar to the text search?

1- how can i build it with tensor flow, nltk… ?
2- how can i return the result of the search as url and article title?
3- how should my data be before processing, in one big json file or multiple ones ?
4-how can i match the text search to the article ?

thx in advanced and plz 🙏🏻 mention me in the reply

[urban kiln]

since if len(s) == numRows:

example: s="abcde", numRows=5
converted should be:
a
b
c
d
e

solution=abcde

damn

i thought ill just simulate each step and then return some "".join(simulated)

ig

yea

doing usaco training too

but im too lazy to think about a way to convert basex num to basey num

nah im pretty bad at coding i think

cant solve most problems

im doing the simulation section rn

i think i am at the lost cow

im able to solve them

but im not fast at solving

i think bronze problems can be solved fast

https://codeforces.com/problemset/problem/1765/E this one is great

1.1k i think

i did 4 contests so far

nah

i think 1.2k can be reached by just doing div3 problem a-c

there isnt much math in easier problems

[haughty mountain]

what rank is 1.1k again?

[urban kiln]

newbie

unrated

[haughty mountain]

ah

2007, though I haven't competed in years

yes

[urban kiln]

i want to reach 1.6 or 1.8 next year

[haughty mountain]

CM isn't amazingly high

depends on how good you are 😛

I was at 1599 after 3 contests, but that was before div3, div4 and everything existed

[urban kiln]

damn

[haughty mountain]

a mix of math, physics and engineering

I did a bunch outside codeforces before

I'm also better at long form contests like the ones hackerrank and codechef had

[urban kiln]

class Solution:
    def convert(self, s: str, numRows: int) -> str:
        if numRows == 1 or numRows == len(s): return s
        simulated = ['' for _ in range(numRows)]
        section = (numRows-1)*2
        for index, i in enumerate(s):
            if index % section < numRows:
                simulated[index % section] += i
            else:
                simulated[(section - (index % section)) % numRows] += i
        return ''.join(simulated)
```section is just the num of chars per section(picture)

but this isnt constant space right?

since it is creating a seperate aarray

[urban kiln]

heuristic ones?

https://leetcode.com/explore/learn/card/array-and-string/202/introduction-to-2d-array/1167/

[crude karma]

What's the command for formatting code in chat? Is it backticks?

[urban kiln]

` 3times on each side

[crude karma]

Thanks 🙂

[urban kiln]

code

[crude karma]

Ok, so I'm working with a coworker's code. It's really something. I'm trying to figure out the Big O on it because it's so slow. Using pseudo code:

function
  for (A)
    for (B)
      for (C)
    for (D)
      for (E)
    for (F)
    for (G)
    while (H)
      for (I)

And we call that function twice on a specific operation.

Is that Big O(2n^9)?

[haughty mountain]

there are some ~week long contests

not heuristic ones

other than practice and pick up new stuff, not much

[haughty mountain]

depends on what the loops are

without knowing ranges for loops you can't say much

and the while loop could be a lot of different things

[crude karma]

Good point, I haven't had to calculate Big O since school. They're not all For (A).

[haughty mountain]

big O is just about counting the number of overall operations and how it scales with parameters

[urban kiln]

were there like 100 problems or extremely hard ones?

[haughty mountain]

you likely have a bunch of parameters, unless all loops are over the same range

[crude karma]

Yeah. In practice, this thing is very slow. But I can't give an accurate Big O based on the loops alone

[haughty mountain]

problems ranging from not too hard to really hard, e.g. an old codechef one:
https://www.codechef.com/JUNE18A/

[urban kiln]

is popping the last element of an array O(n) or O(1)?

[haughty mountain]

last element is O(1)

[urban kiln]

first element is O(n) right?

[haughty mountain]

yes, you need to shift everything after the deleted element

[urban kiln]

oh ok

[humble flower]

what is balanced bst and why do we use it for dynamic lists?

[agile sundial]

we don't

[humble flower]

what do we use it for?

and what is it?

[agile sundial]

sets, maps

reptile already explained it to you

did you not understand the explanation?

[humble flower]

i kinda did but i had some follow up questions and wanted some more clarity

[agile sundial]

ask those then

[humble flower]

but please explain it again because i didnt 100% understand it

[agile sundial]

what parts did you not understand

[humble flower]

basically what it actually does and why we use it

[agile sundial]

what do you mean "what it actually does"

it's used for sets and maps

[humble flower]

what does the algorithm do on a technical level.

what is its purpuse and how does it do that

[agile sundial]

there are multiple ways to implement a BBST. i'm most familiar with the red-black tree. another common one is the AVL tree

[humble flower]

could you explain what those two things are?

[agile sundial]

they're implementations of BBST. i'm not familiar with the AVL tree, but the red black tree essentially "rotates" nodes so that the heights of subtrees are within 1 of each other

[urban kiln]

https://youtu.be/dIuWLtWnkgs

should be english

about red black

[humble flower]

can someone please explain the very basics of balanced bst to me?

[tepid pivot]

Am looking for a study buddy, anyone interested?

[urban kiln]

https://leetcode.com/problems/base-7/

what problems were greed<?

forgot it

damn

i hate question which are edging me

what type of questions are being used in university exams?

are uni exams just some leetcode like questions?

do you have to describe how you solved it and explain it

or is it just like some leetcode weekly contest where you have to answer 4 questions

[urban kiln]

how do i convertz a num to base n where n>10

```py
def to_base(num: int, base: int):
if num < base: return str(num)
return to_base(num // base) + str(num % base)

this works for base<10

[narrow mirage]

str(num % base) needs to be refactored: every num possible between 0 and base-1 (both inclusive) should have a unique character to represent them

you could for example, store a list of those (in base 16 for example:

digits = ['0', '1', ... , '9', 'A', 'B', 'C', 'D', 'E', 'F']

and then you modify str(num%base) by something like digits[num%base]
(I let you code it at your best convenience)

[urban kiln]

ah

yea ty

yea it worked

ty

digits = ['0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J']  # (first 20 digits)
def to_base(num):
    if base == 10:
        return num
    if num < base: return digits[num]
    return to_base(num // base) + digits[num % base]

base isnt given as a parameter because it is read from a file a few lines above

base = int(fin.read())

[urban kiln]

@cobalt mirage https://leetcode.com/problems/keys-and-rooms/ this one is good

it was the daily a few days ago and it was fun solving it

oh

class Solution:
    def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:
        visited = set()
        keys = set([0])
        while keys:
            temp = []
            for i in keys:
                temp.append(i)
                visited.add(i)
            for i in temp:
                for o in rooms[i]:
                    if o not in visited:
                        keys.add(o)
                keys.remove(i)
        return len(visited) == len(rooms)

a lot of loops

yea

simulation

simulating everything

i am not good at graph problems

and simulation worked

0 <= rooms[i].length <= 1000

nvm

time was pretty good

O(nm)?

n is number of rooms

[quasi reef]

anyone have time to explain to me the logic behind a codeforce problem?

[urban kiln]

and m the num of keys in a room

idk

probably

https://leetcode.com/problems/climbing-stairs/

this one is easy

yea

bu mean the new problem i send?

[urban kiln]

fr easy

get a paper and write down the first 5 stairs with possible ways

you will notice a pattern

from functools import cache

class Solution:
    @cache
    def climbStairs(self, n: int) -> int:
        if n in (1, 2, 3): 
            return n
        return self.climbStairs(n-1) + self.climbStairs(n-2)```

[urban kiln]

just fibonacci sequence

i hate making my own cache#

im too lazy#

idk

because i did alot of easies to understand linked lists and basic data structures

and i started dsa by doing leetcode

and i wasnt able to solve medium problems

have you solved todays daily?

i think

icpc was the university competition right>?

uni version of ioi i think

im too bad for ioi

i would have to practice way more

and for a long time

whats quant interview?

first round tho

14/15 points

i did

?

[urban kiln]

ill also do it later

[real oriole]

I was stuck with a problem for some time, thought it could be useful to ask here ,

Would there be an ideal way to find the minimal set of Inputs (I's) for which every alphabet (A,b,c,d... it can be more than 26) is present.
for example for I0 + I2 + I3, alphabets A,B,C,D,E,F, are all present as 1 atleast once so I0I2I3 is a "minimal set". Im probably not doing a good job at explaining it but thanks for the time in advance.

edit: apparently this is called a set cover problem which is NP-hard

[haughty mountain]

yeah, that's minimal set cover

[real oriole]

soo how would you recommend approaching the problem, is it more of a machine learning problem? sorry if I''m asking stupid questions 😅

[haughty mountain]

I would check whether your underlying problem is really this problem or not

if it is really this problem we don't know of good optimal approaches

how large is your input?

for small enough input you could just brute force it

[ionic locust]

How can I check if in an iterable, any object contains a specific attribute in O(1)? i.e how can I check if this list of objects contains an object with id of 1 in O(1)?

I know I can use any but that still goes through the whole list whereas ids should be unique

from dataclasses import dataclass

@dataclass
class Person:
    age: int
    id: int
    name: str

john = Person(10, 1, 'john')
brad = Person(11, 2, 'Brad')

people = [john, brad]

I assume using a dictionary with the id as the key is the best bet but I've already got my schema like this and don't really wanna change it if I don't have to

[real oriole]

there might be 2^20 ish rows and thousands of columns

[haughty mountain]

ouch

if you want to solve that optimally you'll have a bad time

[weak moat]

Hi, anyone happy to discuss sorting points by polar angle?
I've got a way to do it using a key function & the atan2 method, but would it be more efficient to use a custom comparison function (to avoid using a trig function) and then use the cmp_to_key function

[vocal gorge]

Like, by comparing their y/x ratios? That might work, but I wouldn't be sure if it actually would be faster, since math functions are written in C, so the time for calculating the arctangent is likely negligible compared to the rest of the code.

[humble flower]

hello! i was learning about vEB trees when i found this:
u^(1/2) - site array v.cluster[0], ....., v.cluster[u^(1/2) - 1]

Im confused as to why v.cluster[u^(1/2) - 1] is the maximum number the tree can store. someone please explain this to me

[haughty mountain]

if it wasn't python I would for sure recommend doing the non trig way, not only because of performance but because of accuracy

one non-trig compator is to get the quadrant for both points, if they are different then we can immediately tell which is smaller, if they are the same compute the 2d cross product of the vectors and check the sign

[solemn moss]

i'd say O(nk/2)

each loop is len(days_of_week) and 2 is subtracted from k each time

ahh no, it's O(infinity), this code won't stop
did you run that code?

[sage flame]

it does stop

[sage flame]

there's a return statement

[solemn moss]

Your K never changes

[sage flame]

look at the first if condition

it does

[solemn moss]

it can't work

found is False always

can you send that code?

[sage flame]

it turns into true

in the second condition

[solemn moss]

but that condition is always False

because days_of_week[i] is a string

and can't be equal to 5

[sage flame]

it's not equal to 5

it's equal to S

[solemn moss]

ahh

lol

sorry

[solemn moss]

so this was right ^

no

ok.. nevermind

[sage flame]

so it's O(n)

[solemn moss]

It's O(nk) - because we should take worst case
When S is "Sun" and K substracts by 1 on each loop

[sage flame]

so it's quadractic time complexity

[solemn moss]

ye

[sage flame]

you think it's possible to make it linear or not

[solemn moss]

What are you trying to achieve?

This looks weird

[sage flame]

like is this the most effecient way to do it

[solemn moss]

No, i mean, can you tell in words what you want to do? I don't really understand what you try to get here

[sage flame]

given days of the week when entering a number return that day

so enter wed, 2 return fri

[solemn moss]

ye, i see

So, i am really sorry, i was wrong again and your code was O(n + k). 🙂
But you can do it better if you get k % 7 - result will be same, but faster on big k - O(n + k % 7), which is just n really

Also it can be written much simpler :

days_of_week = ["Mon", "Tue", "Wed", "Thu", "Fri", "Sat", "Sun"]
return days_of_week[(days_of_week.index(S) + K) % 7]

[sage flame]

can you explain how it's n + k and not n*k

[solemn moss]

First your code goes until it finds S, in worst case it will be last element, n iterations
Then it substracts 1 from k until it become 0, k steps
Other operations are just coefficients and they doesn't matter
Sum: n + k

[fiery cosmos]

        Hash = self.KeyHash(Key)
        BuckLinkedList = self.Buck[Hash]
        self.StoredKeys[Key] = Hash
        NewNode = BuckLinkedList.ReturnSearch({Key:Value})

        if NewNode == None:
            BuckLinkedList.append({Key:Value})

        else:
            NewNode.StoredValue.StoredValue = Value ```

For some reason, when adding to a hashtable, the key is added recursively for infinity. Can anyone see where this is happening

all the same key

[vocal gorge]

Look at the memory addresses - the issue seems to be that your node's successor is the node itself. So you only have one node, which is its own child.

[fiery cosmos]

ight thanks

[fiery cosmos]

Fun fact: [x^x]' = x^x

[vocal gorge]

you mean, like, derivative? that's false, the derivative is x^x (ln(x)+1).

[urban kiln]

https://cses.fi/problemset/task/1072
for problems like the one above, am i supposed to find the formula myself or can i just look the formula up

because i am sure that i wouldve never been able to come up with something like

and if i am not supposed to look such formulas up, how am i supposed to find one myself

[clever fable]

you are supposed to come up with it on your own

and it shouldn't be that hard

just a bit of casework

[opal oriole]

Experience with counting problems. Start by asking yourself a more simple version of the same problem and try solving that.

(Maybe with pawns and if that is complex choose an even more simple question until you can answer it)

(Also remove / relax constraints. What if you don't care if the knights are attacking each other or not?)

[midnight sorrel]

Hi, I have a task:

Where e is number of opponents that are being eliminated in the round
c is number of remaining opponents. 

What algorithm should I use?

[ivory yacht]

Well, that’s an interesting puzzle. For trying to solve these sorts of things, there’d be two opposite approaches - solving it analytically, by figuring out a formula to give the answer, or by getting the computer to try lots of solutions until you find the best. Now the latter is going to become prohibitively expensive quickly since this looks to increase the number of solutions factorially. I would though suggest trying to solve the first few K values that way, see if you can find a pattern. Or try and derive a formula, or maybe even a partial one.

[lean walrus]

I think it can be solved in O(number of opponents) using dynamic programming

[languid nymph]

is the number of remaining opponents meant before or after elimination?

[last fulcrum]

Hi. I have a question. I have this code

    for enum in enum_fields_list:
        for key, value in master_json_dict.items():
            if enum in value:
                print(key, value)
            else:
                print("Not in the master_json")

Currently, it's very inefficient and I'm looking to improve it. I got some few ideas which are not exactly meeting my requirements but I'll post them below.

So for the code above, the issue is that the enum_fields_list list has about 1752 items. The master_json_dict has also about 27 keys and within each key there is a list with an average of 200 items. This makes the for loop very very inefficient and cause the code to be slow. Double checking also occurs.

My first idea was to do this.

flattened_dict_values = chain.from_iterable(master_json_dict.values())

Which takes all the dicts values and put them into one list so I don't have to have to have an exponent loop. But the problem with this approach for me is that I also need the key of each item that is found. Flattening them does not give me access to the key's but only to the values.

Any help appreciated.

[midnight sorrel]

Before elimination

[lean walrus]

Your approach is not changing asymptotic at all

[languid nymph]

Wouldn't the if enum in value: be O(1) if value was a dictionary instead of array? (I am not good at this so don't take this as a serious answer. I am just wondering)

[lean walrus]

Yes, but value is a list

[last fulcrum]

Really? I think it reduces it because I don't have to check the list diff list every time. I just need to do one in.

for enum in enum_fields_list:
  if enum in flattened_dict_values:
      print('yes')

[last fulcrum]

Yea as denball said value is a list

[lean walrus]

You still need O(enums*list length) time on second line

[last fulcrum]

Ok that makes sense. But then at least the double, triple etc checking would be removed since I have only one list.

[lean walrus]

You can achieve O(1) by using other data structure

[last fulcrum]

Can you name it?

[lean walrus]

No, you have to figure it out for yourself.

[last fulcrum]

Ah. This is for work not an assignment.
Anyway that's fine. Thanks for the help