Count number of trailing zeros in Binary representation of a number using Bitset
Last Updated :
07 May, 2025
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 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 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# 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
<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>
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;
}
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 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))
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));
}
}
}
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));
Time Complexity: O(log N), where N is the value of the input number n.
Auxiliary Space: O(1)
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