Kattis 'Digits' 问题：为什么我的第一组代码超过了时间限制，但我的第二组没有？

Question

I am currently doing a problem on Kattis ( https://open.kattis.com/problems/digits ).

The task description is as follows:

Given any number x0, define a sequence using the following recurrence xi+1=the number of digits in the decimal representation of xi Your task is to determine the smallest positive i such that xi=xi−1.

I came up with code A initially, but after code A failed to meet the time limit, I tinkered around with the code slightly to reach code B.

Sample input

42
5
END

Sample output

3
2

Code A:

#time limit exceeded
def recurrence_index(prev):
    counter = 1
    while True:
        current = len(str(prev))
        if current == prev:
            return counter
        else:
            counter += 1
            prev = current
    

while True:
    initial = input().strip()
    if initial == 'END':
        break
    initial = int(initial)
    print(recurrence_index(initial))

Code B:

def recurrence_index(prev):
    counter = 2
    while True:
        current = len(str(prev))
        if current == prev:
            return counter
        else:
            counter += 1
            prev = current
    

while True:
    initial = input().strip()
    if initial == 'END':
        break
    l = len(initial)
    if l == 1 and int(initial) == 1:
        print(1)
    else:
        print(recurrence_index(l))

Unfortunately, I am not quite sure why Code B is faster than Code A. It seems that I have arrived at the right answer by luck. If anyone can help explain why code B is faster than code A, your help will be greatly appreciated. Thank you!

Answer 1

In the first code, int(initial) is expensive on huge huge integers, and that effort is thrown away because you eventually just call str() on it, which could also be expensive.

In the second code, you just directly call len(initial) , so you are only ever calling str() on small-ish integers (<= 100ish), and then calling len() on those strings, which are even shorter.