Approach: Count factors of 5

def trailingZeroes(n):
    count = 0
    while n > 0:
        n //= 5
        count += n
    return count

Complexity: O(log n) time, O(1) space.

Full Solution — all languages

class Solution:
    def trailingZeroes(self, n):
        count = 0
        while n > 0:
            n //= 5
            count += n
        return count

import java.util.*;
class Solution {
    public int trailingZeroes(int n) {
        int count = 0;
        while (n > 0) {
            n /= 5;
            count += n;
        }
        return count;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int trailingZeroes(int n) {
        int count = 0;
        while (n > 0) {
            n /= 5;
            count += n;
        }
        return count;
    }
};

/**
 * @param {number} n
 * @return {number}
 */ var trailingZeroes = function(n) {
    let count = 0;
    while (n > 0) {
        n = Math.floor(n / 5);
        count += n;
    }
    return count;
};

Factorial Trailing Zeroes

Example 1

Example 2

Example 3

Constraints