🔒 How do i manage to make my code more efficient?

[little vortex]

Here is the problem
Implement pow(x, n), which calculates x raised to the power n (i.e., xn).

Example 1:

Input: x = 2.00000, n = 10
Output: 1024.00000

The code seems to be working but for input such as x =
0.00001
n =
2147483647
I get time exceedded

n = 2.00000
x = 2147483647
if x < 0:
    n = 1 / n
    x = x * (-1)
f=n
h=0
for i in range(x-1):
    n = n*n
    x=x//2
    if x%2!=0:
        h += 1
    if x == 1:
        break

    if x == 0:
        break
for i in range(h):
    n = n*f
h=float(h)
print(n)

This is my second attemp but it doesnt seem to be any better

[verbal gull]

@little vortex

Python help channel opened

Remember to:

• Ask your Python question, not if you can ask or if there's an expert who can help.
• Show a code sample as text (rather than a screenshot) and the error message, if you've got one.
• Explain what you expect to happen and what actually happens.

:warning: Do not pip install anything that isn't related to your question, especially if asked to over DMs.

Closes after a period of inactivity, or when you send !close.

[ivory sorrel]

If I'm understanding correctly you just need to calculate the exponent of a number right so why don't you just use the exponent operator (**) itself?

Like x ** n

Or I'm understanding it wrong

[little vortex]

becouse the task is not to use it

[ivory sorrel]

Oh

[little vortex]

that's the difficulty

[ivory sorrel]

Well then you can just make a for loop to loop n number of times and multiply it to the original number like so-

def pow(base, pow):
    ans = 1
    for _ in range(pow):
        ans *= base
    return ans

[little vortex]

how would it work negative pow?

[ivory sorrel]

hm

[little vortex]

it seems to be pretty the same but without considering the cases

it works but it's not efficient enough to pass the tests

[ivory sorrel]

def pow(base, pow):
    ans = 1
    for _ in range(pow):
        ans *= base
    if pow < 0:
        return 1/ans
    return ans

This should work

I think

[buoyant crest]

This was on leetcode right

[little vortex]

yes

but leetcode accpet the easy way to write x**n

but i want to find diffrent way

[little vortex]

How range works with negative numbers?

[ivory sorrel]

Oh mb

def pow(base, pow):
    ans = 1
    for _ in range(abs(pow)):
        ans *= base
    if pow < 0:
        return 1/ans
    return ans

This has to work now

!e

def pow(base, pow):
    ans = 1
    for _ in range(abs(pow)):
        ans *= base
    if pow < 0:
        return 1/ans
    return ans
print(pow(2, -1))

[verbal gull]

@ivory sorrel :white_check_mark: Your 3.12 eval job has completed with return code 0.

0.5

[ivory sorrel]

Well it works

The abs func just makes the number to positive. And whenever you have a negative exponent it's just basically 1 / b^n (b is base and n is exponent)

So we're just doing 1/ans only if the power was passed as negative

[thorny valley]

It works, but it might not pass the time test

[little vortex]

exactly

[thorny valley]

I'd say python is just straight up too slow for it

[ivory sorrel]

That's literally the challenge

that's the point of this question

[little vortex]

am i right that it takes more time tto work with fractions than with absolute numbers?

[ivory sorrel]

yeah

ig

[little vortex]

so in my second attemp i tried to cut the number of opeerations

but i think i did sth wrong

since 2^4 == (2^2)^2

[thorny valley]

0.00001**2147483647 took 64 seconds with the custom algo (and it gave 0.0 as an answer because float precision)

Using the language implementation took 0.02 seconds, so 320 times faster (it also gave a 0.0 answer, but still)

[thorny valley]

yeah, that's probably the way to go

what did you do?

[little vortex]

n = 2.00000
x = 2147483647
if x < 0:
    n = 1 / n
    x = x * (-1)
f=n
h=0
for i in range(x-1):
    n = n*n
    x=x//2
    if x%2!=0:
        h += 1
    if x == 1:
        break

    if x == 0:
        break
for i in range(h):
    n = n*f
h=float(h)
print(n)

[verbal gull]

Hey @little vortex!

It looks like you pasted Python code without syntax highlighting.

Please use syntax highlighting to improve the legibility of your code and make it easier for us to help you.

To do this, use the following method:
```py
print('Hello, world!')
```

This will result in the following:

print('Hello, world!')```
You can **edit your original message** to correct your code block.

[ivory sorrel]

I'm not sure but you might have to use bitwise operators maybe and do it but I have no experience in it

[thorny valley]

what's the reason for the if x%2!=0: h += 1?

[little vortex]

let suppose the given power is 33 then the number is(( x^16)^2)*x

i could consider division with every prime number also

[thorny valley]

wait, the question states that you have to do x^n, but aren't you doing n^x in this example?

because you are multiplying n and dividing x

[little vortex]

yes

it was a test in pycharm

[thorny valley]

oh ok

[little vortex]

not the solution wasn't confident it even works

[thorny valley]

i think it's close to working

    if x%2!=0:
        h += 1``` The if statement does nothing because you are dividing with //

[little vortex]

it returned infinity when i replaced 0.00001 with 2

n = 2
x = 33
if x < 0:
    n = 1 / n
    x = x * (-1)
f=n
h=0
for i in range(x-1):
    n = n*n

    if x%2!=0:
        h += 1
    x = x // 2
    if x == 1:
        break

    if x == 0:
        break
for i in range(h):
    n = n*f
print(h)
h=float(h)
print(n)

[thorny valley]

!e i think this should work:

x = 2
n = 33

while n != 1:
  if n %2!=0:
    n -=1
    x *=n
  else:
    n /=2
    x*=x
print(x)```

nope

[little vortex]

is there a way to replace rapidly x and n in pycharm?

[thorny valley]

you can do ctrl+D to select all occurences

then change them to something else

[little vortex]

i will try to reproduce the code to leetcode and see the results

[thorny valley]

im don't have the capacity to figure out the algo for this rn, but you're kinda close i think

[little vortex]

you get my idea and what my code does right?

[thorny valley]

yeah

i got a working solution

do you want me to send it?

||py x = 2 n = 33 original_x = x extra_mul = 0 while n != 1: if n %2!=0: n -=1 extra_mul += 1 else: n /=2 x*=x for i in range(extra_mul): x *= original_x print(x)||

[little vortex]

yeah i am about to try one more time and then compare

[thorny valley]

it takes 11 sec for 2**2147483647

0.02 for 0.00001**2147483647

[thorny valley]

PYTHON TAKES 17 SEC LOL this is faster for big numbers i guess

oh wait but they aren't the same result

[little vortex]

so it isn't working?

[thorny valley]

I think so, i'm running it again to check the difference

for 2**33 it works

it takes so long to print it wtf im gonna print just the diff between them

It's currently converting the diff to a string, in order to print it, when it's done I'll send it here

ok so it starts breaking at 2**34

!e ```py
x = 2
n = 34
original_x = x
extra_mul = 0
while n != 1:
if n %2!=0:
n -=1
extra_mul += 1
else:
n /=2
x*=x

for i in range(extra_mul):
x *= original_x
other = 2**34
print(x==other)
print(x)
print(f"{other}")
print(f"{other - x}")```

[verbal gull]

@thorny valley :white_check_mark: Your 3.12 eval job has completed with return code 0.

001 | False
002 | 8589934592
003 | 17179869184
004 | 8589934592

[little vortex]

damn

i think i don't see the better solution

right now

[thorny valley]

I'm honestly not sure why it starts breaking, (why would it work for everything before that???)

lemme try something

[little vortex]

I don't see why it wouldn't mathematicly work so i assume it must be sth with code

n = 2
x = 10
if x < 0:
    n = 1 / n
    x = x * (-1)
f=n
h=0
for i in range(x-1):
    n = n*n

    if x%2!=0:
        h += 1
    x = x // 2
    if x == 1:
        break

for i in range(h):
    n = n*f
print(h)
h=float(h)
print(n)

whis code returns the same for 9 and 10

[thorny valley]

yeah, that's the weird part

[little vortex]

okey i see what's the problem

but i don't see the solution

[thorny valley]

yeah it's the extra_mul, because we're not considering x when addingto the power

[little vortex]

it's the fact that 10 //2=4 and 4*2=8

so it lack 1 extra multiplication

[little vortex]

this

[thorny valley]

yeah i have a working solution

add the current x to a list when we should add to the multiplication in the end, and then in the end loop over the list and multiply x by each item

let me just check it works for 2**2147483647

yup it does

It takes 33 sec though

x =2
n = 2147483647
original_x = x
muls = []
while n != 1:
  if n %2!=0:
    n -=1
    muls.append(x)
  else:
    n /=2
    x*=x
for i in muls:
    x *= i```

Wait you can just *= mul by the current x, no list needed

[little vortex]

deam that's crazy u have detected that

[thorny valley]

!e py x =2 n = 2147483647 original_x = x mul =1 while n != 1: if n %2!=0: n -=1 mul*=x else: n /=2 x*=x x*=mul print(x)

[verbal gull]

@thorny valley :warning: Your 3.12 eval job timed out or ran out of memory.

[No output]

[thorny valley]

Crap. I'm not at my computer right now, but this should work

[little vortex]

thank i am about ot try in next day this one made my brain freeze

this seemed to be the easy one

[thorny valley]

The simplest ones often have some detail (like optimizing the time) that fuck you up. The other ones once you figure out the trick are easier imo

[verbal gull]

Python help channel closed

This help channel has been closed and it's no longer possible to send messages here. If your question wasn't answered, feel free to create a new post in #1035199133436354600. To maximize your chances of getting a response, check out this guide on asking good questions.