Java Program to Count trailing zeroes in factorial of a number

Given an integer n, write a function that returns count of trailing zeroes in n!.  Examples :

Input: n = 5
Output: 1 
Factorial of 5 is 120 which has one trailing 0.

Input: n = 20
Output: 4
Factorial of 20 is 2432902008176640000 which has
4 trailing zeroes.

Input: n = 100
Output: 24

Trailing 0s in n! = Count of 5s in prime factors of n!
                  = floor(n/5) + floor(n/25) + floor(n/125) + ....

Java

// Java program to count
// trailing 0s in n!
import java.io.*;

class GFG {
    // Function to return trailing
    // 0s in factorial of n
    static int findTrailingZeros(int n)
    {
        // Initialize result
        int count = 0;

        // Keep dividing n by powers
        // of 5 and update count
        for (int i = 5; n / i >= 1; i *= 5)
            count += n / i;

        return count;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int n = 100;
        System.out.println("Count of trailing 0s in "
                          + n + "! is " + findTrailingZeros(n));
    }
}

// This code is contributed by Pramod Kumar

Output:

Count of trailing 0s in 100! is 24

Time Complexity:  O(log₅n)

Auxiliary Space: O(1)

Please refer complete article on Count trailing zeroes in factorial of a number for more details!