Previous_multiple_of_three.md

CodeWars Python Solutions

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Previous multiple of three

Given a positive integer n: 0 < n < 1e6, remove the last digit until you're left with a number that is a multiple of three.

Return n if the input is already a multiple of three, and return null/nil/None if no such number exists.

Examples

1      => null
25     => null
36     => 36
1244   => 12
952406 => 9

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Given Code

def prev_mult_of_three(n):
    pass

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Solution 1

def prev_mult_of_three(n):
    if n % 3 == 0:
        resp = n
    elif n > 3:
        try:
            resp = prev_mult_of_three(int("".join([i for i in str(n)[:-1]])))
        except:
            resp = None
    else:
        resp = None

    return resp

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Solution 2

def prev_mult_of_three(n):
    while n % 3:
        n //= 10
    return n or None

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Solution 3

def prev_mult_of_three(n):
    return prev_mult_of_three(n//10) if n%3 else n or None

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