Number of trailing zeros in N * (N – 2) * (N – 4)*….

Given an integer N, the task is to find the number of trailing zeros in the decimal notation of f(N) where f(N) = 1 if N < 2 and f(N) = N * f(N – 2) if N ≥ 2

Examples:

Input: N = 12 
Output: 1 
f(12) = 12 * 10 * 8 * 6 * 4 * 2 = 46080

Input: N = 7 
Output: 0 

Approach: The number of trailing zeros when f(N) is expressed in decimal notation is the number of times f(N) is divisible by 2 and the number of times f(N) is divisible by 5. There are two cases:  

1. When N is odd then f(N) is the product of some odd numbers, so it does not break at 2. So the answer is always 0.
2. When N is even then f(N) can be represented as 2 (1 * 2 * 3 * …. * N/2). The number of times f(N) is divisible by 2 is greater than the number of times divisible by 5, so only consider the number of times divisible by 5. Now, this problem is similar to count trailing zeroes in factorial of a number.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count of
// trailing 0s in the given function
int findTrailingZeros(int n)
{
    // If n is odd
    if (n & 1)
        return 0;

    // If n is even
    else {
        int ans = 0;

        // Find the trailing zeros
        // in n/2 factorial
        n /= 2;
        while (n) {
            ans += n / 5;
            n /= 5;
        }

        // Return the required answer
        return ans;
    }
}

// Driver code
int main()
{
    int n = 12;

    cout << findTrailingZeros(n);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

    // Function to return the count of
    // trailing 0s in the given function
    static int findTrailingZeros(int n)
    {
        // If n is odd
        if ((n & 1) == 1)
            return 0;

        // If n is even
        else
        {
            int ans = 0;

            // Find the trailing zeros
            // in n/2 factorial
            n /= 2;
            while (n != 0)
            {
                ans += n / 5;
                n /= 5;
            }

            // Return the required answer
            return ans;
        }
    }

    // Driver code
    public static void main (String[] args)
    {
        int n = 12;

        System.out.println(findTrailingZeros(n));
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach

# Function to return the count of
# trailing 0s in the given function
def findTrailingZeros(n):

    # If n is odd
    if (n & 1):
        return 0

    # If n is even
    else:
        ans = 0

        # Find the trailing zeros
        # in n/2 factorial
        n //= 2
        while (n):
            ans += n // 5
            n //= 5

        # Return the required answer
        return ans

# Driver code

n = 12

print(findTrailingZeros(n))

# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the count of
    // trailing 0s in the given function
    static int findTrailingZeros(int n)
    {
        // If n is odd
        if ((n & 1) == 1)
            return 0;

        // If n is even
        else
        {
            int ans = 0;

            // Find the trailing zeros
            // in n/2 factorial
            n /= 2;
            while (n != 0)
            {
                ans += n / 5;
                n /= 5;
            }

            // Return the required answer
            return ans;
        }
    }

    // Driver code
    public static void Main(String[] args)
    {
        int n = 12;

        Console.WriteLine(findTrailingZeros(n));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript implementation of the approach

// Function to return the count of
// trailing 0s in the given function
function findTrailingZeros(n)
{

    // If n is odd
    if (n & 1)
        return 0;

    // If n is even
    else
    {
        let ans = 0;

        // Find the trailing zeros
        // in n/2 factorial
        n = parseInt(n / 2);

        while (n)
        {
            ans += parseInt(n / 5);
            n = parseInt(n / 5);
        }

        // Return the required answer
        return ans;
    }
}

// Driver code
let n = 12;

document.write(findTrailingZeros(n));

// This code is contributed by subhammahato348

</script>

Output:

1

Time Complexity: O(log₅n)

Auxiliary Space: O(1)