🔒 someone guide me in my coding solution

[lost mauve]

def lucky_seven(n):
    '''
    Assume n is an integer. A number is considered lucky if it 
    has more 7 digits than 0 digits. How many lucky numbers are 
    there between 1-n inclusive?
    For example, 771203 is lucky, but 123 and 207 are not.
    '''
    seven_count = 0
    zero_count = 0
    else_count = 0
    for num in range(n):
        if num == 7:
            seven_count += 1
        elif num == 0:
            zero_count += 1
        else:
            else_count += 0
            
    
    if seven_count > zero_count:
        return(seven_count)
    else:
        return("0")
print(lucky_seven(771203))

[alpine matrix]

@lost mauve

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[lost mauve]

im not 100% sure i know what the question is asking

[raw scroll]

Suppose you want to calculate lucky_seven(100). There are 100 numbers between 1 and 100 inclusive: 1, 2, 3, ..., 100.
Of these, it so happens that there are 18 numbers with more sevens than zeros: [7, 17, 27, 37, 47, 57, 67, 71, 72, 73, 74, 75, 76, 77, 78, 79, 87, 97].
So lucky_seven(100) is 18.

[lost mauve]

oh

i get it now

what are the steps to code

it seems hard to compare 7's with 0's for each number

[raw scroll]

It's easy if you convert the number to a string first.

(if I encountered this problem on a contest or something, I'd expect the inputs to be, like, 10000000 and hence one needs to write a solution that doesn't need to consider each number in order to calculate how many there'll be. But I don't know where you got this from, so.)

[lost mauve]

how can i fix the middle of the code

[mortal wedge]

I think you might have meant to use if instead of for

you should be checking though if num_str == "7"

and the same for 0

I don't know why you're checking 6 though

actually that still wouldn't quite be right. Consider using str.count

!d str.count

[alpine matrix]

str.count

str.count(sub[, start[, end]])```
Return the number of non-overlapping occurrences of substring *sub* in the range [*start*, *end*]. Optional arguments *start* and *end* are interpreted as in slice notation.

If *sub* is empty, returns the number of empty strings between characters which is the length of the string plus one.

[lost mauve]

i figured out how to do with with str.count

but i want to see if theres a way to do it with code

i dont think theres a easy way of doing it without using .count

[languid willow]

do you see how you would do this with a for loop?

[lost mauve]

yea i used one

[languid willow]

how does that code look?

[lost mauve]

oh nvm i though this was considered using a for loop

[languid willow]

did you find a solution without using .count()?

[lost mauve]

nope

[languid willow]

you want to see one?

or just have hints?

[lost mauve]

better to see

[languid willow]

!e

n = 771204
seven = 0
zero = 0
for num in str(n):
    if num == "7":
        seven += 1
    elif num == "0":
        zero += 1

print(f"{seven = }")
print(f"{zero = }")

[alpine matrix]

@languid willow :white_check_mark: Your 3.12 eval job has completed with return code 0.

001 | seven = 2
002 | zero = 1

[languid willow]

you run through all the characters of the number one by one and compare them with another string of one character and if it matches you increment that counter

[lost mauve]

hmm

i see

i was thinking something very similar to that

are you able to help with this

def color_mixer(s):
    '''
    Assume that s is a string containing only the characters 
    'r','g','b'. Create and return a new string whose characters 
    are a mix of adjacent colors as follows: 
    'rg' or 'gr' mixes to 'y' (yellow) 
    'gb' or 'bg' mixes to 'c' (cyan) 
    'rb' or 'br' mixes to 'm' (magenta) 
    If adjacent letters are the same, the color stays the same. 
    For example: 
    'rgb' should return 'yc'
    'bgrrgb' should return 'cyryc'
    '''
    string = " "
    for i in range(len(s)):
        if s[i:i+1]  == "rg" or s[i:i+1]  == "gr":
            string = string[1:] + "y"
            return string
        elif s[i:i+1]  == "gb" or s[i:i+1]  =="bg":
            string = string[1:] + "c"
            return string
        elif s[i:i+1]  == "rb" or s[i:i+1]  =="br":
            string = string[1:] + "m"
            return string
        else:
            string = string[1:]
            return string

print(color_mixer("bgrrgb"))

[languid willow]

when you send code here on discord you can put the two characters py just after the three backticks on the first line and then begin write the code on the next line

you can even edit this code up there to do that after the fact

[lost mauve]

changed

[languid willow]

much nicer to read that way 🙂

have you seen this kind of syntax before?

if s[i:i+1] in ("rg", "gr"):

is will compare i data with any of the elements of that tuple

so it's kind of a or condition, but it's shorter, especially if you have many patterns to test against in the same if statement

put a print(s[i:i+1]) on a new line between the line with the for and the if and see what it is you are getting

[languid willow]

after adding that print statement, look at what is being printed on each line, i'm guessing that is not what you are expecting

after that you might see what needs to be changed to get what you expect being printed on each of those lines

[alpine matrix]

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