LeetCode_Practice

Practice Leetcode Problem

def trailingZeroes(self, n: int) -> int:
        
    # Calculate n!
    n_factorial = 1
    for i in range(2, n + 1):
        n_factorial *= i
    
    # Count how many 0's are on the end.
    zero_count = 0
    while n_factorial % 10 == 0:
        zero_count += 1
        n_factorial //= 10
        
    return zero_count

def trailingZeroes(self, n: int) -> int:
        
    zero_count = 0
    for i in range(5, n + 1, 5):
        current = i
        while current % 5 == 0:
            zero_count += 1
            current //= 5

    return zero_count

def trailingZeroes(self, n: int) -> int:
        
    zero_count = 0
    for i in range(5, n + 1, 5):
        power_of_5 = 5
        while i % power_of_5 == 0:
            zero_count += 1
            power_of_5 *= 5

    return zero_count

def trailingZeroes(self, n: int) -> int:
    zero_count = 0
    current_multiple = 5
    while n >= current_multiple:
        zero_count += n // current_multiple
        current_multiple *= 5
    return zero_count

def trailingZeroes(self, n: int) -> int:
    zero_count = 0
    while n > 0:
        n //= 5
        zero_count += n
    return zero_count

Complexity Analysis

Time complexity : O(\log , n)O(logn).

In this approach, we divide nn by each power of 55. By definition, there are \log_5nlog 5 ​ n powers of 55 less-than-or-equal-to nn. Because the multiplications and divisions are within the 32-bit integer range, we treat these calculations as O(1)O(1). Therefore, we are doing \log_5 n \cdot O(1) = \log , nlog 5 ​ n⋅O(1)=logn operations (keeping in mind that \loglog bases are insignificant in big-oh notation).

Space complexity : O(1)O(1).

We use only a fixed number of integer variables, therefore the space complexity is O(1)O(1).