## Factorial Trailing Zeroes

Given an integer

n, return the number of trailing zeroes inn!.

Note:Your solution should be in logarithmic time complexity.

Credits:

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

The factor of 10 is 2 and 5. Given N!, the factor of 5 is always less than 2. So we can just count the number of factor 5 in the factorial of N.

Be careful of the overflow issue.

class Solution { public: int trailingZeroes(int n) { int ans = 0; long factor = 5; // BUG overflow while(n >= factor){ ans += n / factor; factor = factor * 5; } return ans; } };