172. Factorial Trailing Zeroes

172. Factorial Trailing Zeroes

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Total Accepted: 58597  Total Submissions: 178269  Difficulty: Easy

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

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class Solution {
public:
    int trailingZeroes(int n) {
        int ans=0;
        while(n)
        {
            ans+=n/5;
            n/=5;
        }
        return ans;
    }
};