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!