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Was asked in an interview today -

You are given an integer (in finite range [-N, N]) and you wish to maximize the number by inserting somewhere the digit '5'.

My solution - the idea here is going from left to right and checking if the digit is bigger than 5 or not. The first time you reach one that isn't you insert the 5 there. With the exception of negative numbers where you check and do the opposite.

I did it by converting ints to strings and vice versa.

def solution(N):
    array = str(N)
    result = []
    mod = 1
    for i in range(len(array) + 1):
        if i == len(array):
            result.append('5')
            break

        if array[i] == '-':
            mod = -1
            result.append(array[i])
            continue

        if int(array[i]) * mod > 5 * mod:
            result.append(array[i])
        else:
            result.append('5')
            result.append(array[i:])
            break

    return int("".join(result))

So - solution(128) will output 5128; solution(0) will output 50, solution(-74) will output -574 etc.

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Presumably, if array[i] == '-' is meant to check if the number is negative. Having it like you have though doesn't make a lot of sense. If it's in the loop, that suggests that a - could be expected at any point within the number. In reality though, it will only ever occur as the first character in the string. I'd just check N the number with if N < 0 to see if it's negative. You can streamline this check right in the definition of mod:

mod = 1 if N >= 0 else -1

Unfortunately, that slightly complicates adding - to result. The simplest way to deal with it I think is just to change the definition of result to be similar to that of mod:

result = [] if n >= 0 else ['-']

Since the checks are the same though, you could combine then into returning a tuple instead then unpacking it:

mod, result = (1, []) if n >= 0 else (-1, ['-'])

And, array will need to have the - cleaved off if we're doing it before the loop:

mod, result, checked_digits = (1, [], digits) if n >= 0 else (-1, ['-'], digits[1:])

Normally, I don't think a triple variable definition like this should be encouraged, but there's nothing too fancy or complicated going on here to make the line confusing. I think it's fine.


I'd rename array and N:

  • array isn't really a correct or useful description. I think digits would be better.

  • N should be n.


You're iterating over the indices of digits and using digits[i] all over the place even though you don't even need the index in most cases. I'd iterate over the digits themselves, and use enumerate to get the index in the couple cases that you actually need it for:

for i, digit in enumerate(digits):
    . . .

    if int(digit) * mod > 5 * mod:
        result.append(digit)

I'd move the if i == len(array): check out of the loop as well. Really, it's just checking if at no point a break was encountered. That's exactly the use case for for's else clause (yes, for can have an else):

for i, digit in enumerate(checked_digits):
    . . .

else:
    result.append(5)

The else will only execute if no break was encountered. In the case of this code, that means that no digit greater than 5 was found.

Honestly, this is the first time I've ever found a useful case for for...else. It worked out nice here though.


I'd add 5 and '5' as constants at the top. Having the magic strings and numbers floating your code makes your code harder to change latter, and doesn't indicate what the 5 and '5' actually are for:

TARGET_NUM = 5
TARGET_DIGIT = str(TARGET_NUM)


def solution(n):

Or, just make that number a parameter with a default to make it easy to use different target numbers:

def solution(n: int, target_num: int = 5) -> int:
    target_digit = str(target_num)

I also added some type hints in there, because why not?



In the end, this is what I ended up with:

def solution(n: int, target_num: int = 5):
    target_digit = str(target_num)
    digits = str(n)

    mod, result, checked_digits = (1, [], digits) if n >= 0 else (-1, ['-'], digits[1:])

    for i, digit in enumerate(checked_digits):
        if int(digit) * mod > TARGET_NUM * mod:
            result.append(digit)

        else:
            result.append(TARGET_DIGIT)
            result.append(checked_digits[i:])
            break

    else:
        result.append(TARGET_DIGIT)

    return int("".join(result))
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  • \$\begingroup\$ thanks for the input. Never knew about the for-else thingy \$\endgroup\$ – David Refaeli Dec 7 '19 at 12:22

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