Count Number of Digits

Naive:

Time Complexity: O(log₁₀(n))

Space Complexity: O(1)

def countDigits(n:int) -> int:
    int res = 0
    while (x>0):
        x = x//10
        res += 1
    return res

Recursive Way:

Time Complexity: O(log₁₀(n))

Space Complexity: O(log₁₀(n))

def countDigits(n:int) -> int:
    if n//10==0:
        return 1
    return 1 + countDigits(n//10)

Best Way:

Time Complexity: O(log₁₀(n))

Space Complexity: O(log₁₀(n)) — due to string conversion

def countDigits(n: int) -> int:
    return len(str(n))