Count trailing zeroes present in binary representation of a given number using XOR

Given an integer N, the task is to find the number of trailing zeroes in the binary representation of the given number.

Examples:

Input: N = 12
Output: 2
Explanation:
The binary representation of the number 13 is “1100”.
Therefore, there are two trailing zeros in the 12.

Input: N = -56
Output: 3
Explanation:
The binary representation of the number -56 is “001000”.
Therefore, there are 3 trailing zeros present in -56.

Approach: Follow the steps to solve the problem

• The idea is to use the observation that after calculating XOR of N with N – 1, all the set bit of N left to the rightmost set bit, i.e LSB set bit disappears and the rightmost set bit of N becomes the leftmost set bit of N ^ (N – 1).
• Print the count of bits of a number (N ^ (N – 1)) decremented by 1.

Below is the implementation of the above approach:

C++

// C++ implementation
// of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to print count of
// trailing zeroes present in
// binary representation of N
int countTrailingZeroes(int N)
{
    // Count set bits in (N ^ (N - 1))
    int res = log2(N ^ (N - 1));

    // If res < 0, return 0
    return res >= 0 ? res : 0;
}

// Driver Code
int main()
{
    int N = 12;

    // Function call to print the count
    // of trailing zeroes in the binary
    // representation of N
    cout << countTrailingZeroes(N);

    return 0;
}

Java

// Java implementation
// of the above approach
import java.io.*;

class GFG {

    // Function to print count of
    // trailing zeroes present in
    // binary representation of N
    public static int countTrailingZeroes(int N)
    {
        // Stores XOR of N and (N-1)
        int res = N ^ (N - 1);

        // Return count of set bits in res
        return (int)(Math.log(temp)
                     / Math.log(2));
    }

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

        // Function call to print the count
        // of trailing zeroes in the binary
        // representation of N
        System.out.println(
            countTrailingZeroes(N));
    }
}

Python3

# Python3 implementation
# of the above approach
from math import log2

# Function to print count of
# trailing zeroes present in
# binary representation of N
def countTrailingZeroes(N):

    # Count set bits in (N ^ (N - 1))
    res = int(log2(N ^ (N - 1)))

    # If res < 0, return 0
    return res if res >= 0 else 0

# Driver Code
if __name__ == '__main__':
    N = 12

    # Function call to print the count
    # of trailing zeroes in the binary
    # representation of N
    print (countTrailingZeroes(N))

    # This code is contributed by mohit kumar 29.

C#

// C# implementation
// of the above approach
using System;
public class GFG{

  // Function to print count of
  // trailing zeroes present in
  // binary representation of N
  public static int countTrailingZeroes(int N)
  {
    // Stores XOR of N and (N-1)
    int res = (int)Math.Log(N ^ (N - 1), 2.0);

    // Return count of set bits in res
    if(res >= 0)
      return res;
    else
      return 0;
  }

  // Driver Code
  static public void Main ()
  {

    int N = 12;

    // Function call to print the count
    // of trailing zeroes in the binary
    // representation of N
    Console.WriteLine(
      countTrailingZeroes(N));
  }
}

// This code is contributed by Dharanendra L V.

Javascript

<script>

// Javascript implementation
// of the above approach

// Function to print count of
// trailing zeroes present in
// binary representation of N
function countTrailingZeroes(N)
{

    // Count set bits in (N ^ (N - 1))
    let res = parseInt(Math.log(N ^ (N - 1)) /
                       Math.log(2));

    // If res < 0, return 0
    return res >= 0 ? res : 0;
}

// Driver Code
let N = 12;

// Function call to print the count
// of trailing zeroes in the binary
// representation of N
document.write(countTrailingZeroes(N));

// This code is contributed by souravmahato348

</script>

Output:

2

Time Complexity: O(log(N))
Auxiliary Space: O(1)