Description

Write an algorithm which computes the number of trailing zeros in n factorial.

Example

11! = 39916800, so the out should be 2

Challenge

O(log N) time

Answer

    /*
     * @param n: A long integer
     * @return: An integer, denote the number of trailing zeros in n!
     */
    long long trailingZeros(long long n) {
        // write your code here, try to do it without arithmetic operators.
        if(n<5){
            return 0;
        }
        else{
            return n/5 + trailingZeros(n/5);
        }
    }

Tips

This solution is implemented by a recursive method, we can also use a loop method to solve this problem.

    /*
     * @param n: A long integer
     * @return: An integer, denote the number of trailing zeros in n!
     */
    long long trailingZeros(long long n) {
        // write your code here, try to do it without arithmetic operators.
        long long result = 0;
        while ( n > 0)
        {
            result += n/5;
            n /= 5;
        }
        
        return result;
    }