Count number of trailing zeros in Binary representation of a number using Bitset

Given a number. The task is to count the number of Trailing Zero in Binary representation of a number using bitset.

Examples:

Input : N = 16
Output : 4
Binary representation of N is 10000. Therefore,
number of zeroes at the end is 4.
Input : N = 8
Output : 3

Approach: We simply set the number in the bitset and then we iterate from 0 indexes of bitset, as soon as we get 1 we will break the loop because there is no trailing zero after that.

Below is the implementation of above approach: 

C++

// C++ program to count number of trailing zeros
// in Binary representation of a number
// using Bitset

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

// Function to count number of trailing zeros in
// Binary representation of a number
// using Bitset
int CountTrailingZeros(int n)
{
    // declare bitset of 64 bits
    bitset<64> bit;

    // set bitset with the value
    bit |= n;

    int zero = 0;

    for (int i = 0; i < 64; i++) {
        if (bit[i] == 0)
            zero++;
        // if '1' comes then break
        else
            break;
    }

    return zero;
}

// Driver Code
int main()
{
    int n = 4;

    int ans = CountTrailingZeros(n);

    cout << ans << "\n";

    return 0;
}

Java

// Java program to count number of trailing zeros
// in Binary representation of a number
// using Bitset
import java.util.*;
import java.lang.*;
import java.io.*;

class GFG
{

    // Function to count number of trailing zeros in
    // Binary representation of a number
    // using Bitset
    static int CountTrailingZeros(int n)
    {

        String bit = Integer.toBinaryString(n);
        StringBuilder bit1 = new StringBuilder();
        bit1.append(bit);
        bit1=bit1.reverse();
        int zero = 0;

        for (int i = 0; i < 64; i++) {
            if (bit1.charAt(i) == '0')
                zero++;
            // if '1' comes then break
            else
                break;
        }

        return zero;
    }

    // Driver Code
    public static void main(String []args)
    {
        int n = 4;

        int ans = CountTrailingZeros(n);

        System.out.println(ans);
    }
}

// This code is contributed by chitranayal

Python3

# Python3 program to count
# number of trailing zeros in
# Binary representation of a number

# Function to count number of
# trailing zeros in Binary
# representation of a number
def CountTrailingZeros(n):

    # declare bitset of 64 bits
    bit = bin(n)[2:]
    bit = bit[::-1]

    zero = 0;

    for i in range(len(bit)):
        if (bit[i] == '0'):
            zero += 1

        # if '1' comes then break
        else:
            break

    return zero

# Driver Code
n = 4

ans = CountTrailingZeros(n)

print(ans)

# This code is contributed
# by Mohit Kumar

C#

// C# program to count number of trailing zeros
// in Binary representation of a number
// using Bitset
using System;
class GFG
{

  // Function to count number of trailing zeros in
  // Binary representation of a number
  // using Bitset
  static int CountTrailingZeros(int n)
  {
    string bit=Convert.ToString(n, 2);
    char[] charArray = bit.ToCharArray();
    Array.Reverse( charArray );
    string bit1 = new string( charArray );
    int zero = 0;
    for (int i = 0; i < 64; i++)
    {
      if (bit1[i] == '0')
      {
        zero++;
      }

      // if '1' comes then break
      else
      {
        break;
      }
    }
    return zero;
  }

  // Driver Code
  static public void Main ()
  {
    int n = 4;
    int ans = CountTrailingZeros(n);
    Console.WriteLine(ans);
  }
}

// This code is contributed by avanitrachhadiya2155

Javascript

<script>

// JavaScript program to count number of trailing zeros
// in Binary representation of a number
// using Bitset

    // Function to count number of trailing zeros in
    // Binary representation of a number
    // using Bitset
    function CountTrailingZeros(n)
    {
        let bit = n.toString(2);
        let bit1=bit.split("");
        bit1=bit1.reverse();
        let zero = 0;

        for (let i = 0; i < 64; i++) {
            if (bit1[i] == '0')
                zero++;
            // if '1' comes then break
            else
                break;
        }

        return zero;
    }

    // Driver Code
    let n = 4;
    let ans = CountTrailingZeros(n);
    document.write(ans);



// This code is contributed by rag2127

</script>

Output

2

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

Approach: Bitwise Operation

• To count the number of trailing zeroes in the binary representation of a number using bitwise operations.
• We can repeatedly right-shift the number by 1 until the least significant bit (LSB) becomes 1. 
• The number of right shifts performed will give us the count of trailing zeroes.

Below is the implementation of above approach: 

C++

#include <iostream>
using namespace std;

int count_trailing_zeroes(int n) {
    int count = 0;
    while ((n & 1) == 0) {
        count += 1;
        n >>= 1;
    }
    return count;
}

int main() {
    int n1 = 16;
    cout << count_trailing_zeroes(n1) << endl;

    return 0;
}

Java

import java.util.Scanner;

public class GFG {

    // Function to count trailing zeroes in a number
    static int countTrailingZeroes(int n) {
        int count = 0;
        while ((n & 1) == 0) {
            count += 1;
            n >>= 1;
        }
        return count;
    }

    // Driver Code
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        System.out.print("Enter an integer: ");
        int n1 = 16;

        System.out.println("Number of trailing zeroes: " + countTrailingZeroes(n1));
    }
}

Python

# Python program to count
# number of trailing zeros in
# Binary representation of a number


def count_trailing_zeroes(n):
    count = 0
    while n & 1 == 0:
        count += 1
        n >>= 1
    return count


# Driver Code
n1 = 16
print(count_trailing_zeroes(n1))

C#

using System;

namespace CountTrailingZeroes
{
    class Program
    {
         // Function to count trailing zeroes in a number
        static int CountTrailingZeroes(int n)
        {
            int count = 0;

            while ((n & 1) == 0)
            {
                count += 1;
                n >>= 1;
            }

            return count;
        }

        static void Main(string[] args)
        {
            int n1 = 16;
            Console.WriteLine(CountTrailingZeroes(n1));
        }
    }
}

Javascript

function countTrailingZeroes(n) {
    let count = 0;
    while ((n & 1) === 0) {
        count += 1;
        n >>= 1;
    }
    return count;
}

// Example usage
const n1 = 16;
console.log(countTrailingZeroes(n1));

Output

4

Time Complexity: O(log N), where N is the value of the input number n.
Auxiliary Space: O(1)