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Check whether the bit at given position is set or unset

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Given two positive integers n and k. The problem is to check whether the bit at position k from the right in the binary representation of n is set (‘1’) or unset (‘0’). 
Constraints: 1 <= k <= number of bits in the binary representation of n.
Examples: 
 

Input : n = 10, k = 2
Output : Set
(10)10 = (1010)2
The 2nd bit from the right is set.

Input : n = 21, k = 4
Output : Unset


 


Approach#1: Following are the steps:
 

  1. Calculate new_num = (n >> (k – 1)).
  2. if (new_num & 1) == 1 then bit is “Set”, else “Unset”.


 

C++

// C++ implementation to check whether the bit
// at given position is set or unset
#include <bits/stdc++.h>
using namespace std;
 
// function to check whether the bit
// at given position is set or unset
bool bitAtGivenPosSetOrUnset(unsigned int n,
                             unsigned int k)
{
    int new_num = n >> (k - 1);
 
    // if it results to '1' then bit is set,
    // else it results to '0' bit is unset
    return (new_num & 1);
}
 
// Driver program to test above code
int main()
{
    unsigned int n = 10, k = 2;
    if (bitAtGivenPosSetOrUnset(n, k))
        cout << "Set";
    else
        cout << "Unset";
    return 0;
}

                    

Java

// Java program to
// check the set bit
// at kth position
import java.io.*;
 
class GFG {
     
// function to check whether
// the bit at given position
// is set or unset
static int bitAtGivenPosSetOrUnset
                   ( int n, int k)
{
 
    // to shift the kth bit
    // at 1st position
    int new_num = n >> (k - 1);
  
    // Since, last bit is now
    // kth bit, so doing AND with 1
    // will give result.
    return (new_num & 1);
}
    public static void main (String[] args)
    {
         // K and n must be greater than 0
         int n = 10, k = 2;
          
    if (bitAtGivenPosSetOrUnset(n, k)==1)
        System.out.println("Set");
    else
        System.out.println("Unset");
    }
}
 
//This code is contributed by Gitanjali

                    

Python3

# python implementation to check
# whether the bit at given
# position is set or unset
 
import math
#function to check whether the bit
# at given position is set or unset
def bitAtGivenPosSetOrUnset(  n,  k):
     new_num = n >> (k - 1)
  
     #if it results to '1' then bit is set,
     #else it results to '0' bit is unset
     return (new_num & 1)
 
# Driver code
n = 10
k = 2
if (bitAtGivenPosSetOrUnset(n, k)):
     print("Set")
else:
    print("Unset")
 
#This code is contributed by Gitanjali

                    

C#

// C# program to check the set bit
// at kth position
using System;
 
class GFG {
     
    // function to check whether
    // the bit at given position
    // is set or unset
    static int bitAtGivenPosSetOrUnset(
                           int n, int k)
    {
     
        // to shift the kth bit
        // at 1st position
        int new_num = n >> (k - 1);
     
        // Since, last bit is now
        // kth bit, so doing AND with 1
        // will give result.
        return (new_num & 1);
    }
     
    // Driver code
    public static void Main ()
    {
         
        // K and n must be greater
        // than 0
        int n = 10, k = 2;
         
        if (bitAtGivenPosSetOrUnset(n, k)==1)
            Console.Write("Set");
        else
            Console.Write("Unset");
    }
}
 
// This code is contributed by Sam007.

                    

PHP

<?php
// PHP implementation to check whether the bit
// at given position is set or unset
 
// function to check whether the bit
// at given position is set or unset
function bitAtGivenPosSetOrUnset($n, $k)
{
    $new_num = $n >> ($k - 1);
 
    // if it results to '1' then bit is set,
    // else it results to '0' bit is unset
    return ($new_num & 1);
}
 
    // Driver Code
    $n = 10;
    $k = 2;
    if (bitAtGivenPosSetOrUnset($n, $k))
        echo "Set";
    else
        echo "Unset";
         
// This code is contributed by Sam007
?>

                    

Javascript

<script>
 
// javascript program to
// check the set bit
// at kth position
 
// function to check whether
// the bit at given position
// is set or unset
function bitAtGivenPosSetOrUnset
                   (n, k)
{
   
    // to shift the kth bit
    // at 1st position
    let new_num = n >> (k - 1);
    
    // Since, last bit is now
    // kth bit, so doing AND with 1
    // will give result.
    return (new_num & 1);
}
 
// Driver Function
 
         // K and n must be greater than 0
         let n = 10, k = 2;
            
    if (bitAtGivenPosSetOrUnset(n, k)==1)
        document.write("Set");
    else
        document.write("Unset");
     
    // This code is contributed by susmitakundugoaldanga.
</script>

                    

Output: 
 

Set


Time Complexity: O(1)

Auxiliary Space: O(1)

Approach#2: We can use the left shift to solve this problem. Following are the steps:

  • New_num = ( 1 << ( k – 1 ) )
  • If ( num & New_num ) == 1 then bit is set else unset. 

C++

// C++ implementation to check whether the bit
// at given position is set or unset
// Using left shift operator
#include <bits/stdc++.h>
using namespace std;
 
// function Using left shift operator
bool bitAtGivenPosSetOrUnset(unsigned int n,
                             unsigned int k)
{
    int New_num = 1 << (k - 1);
 
// Returning result
    return (New_num & n);
}
 
// Driver program to test above code
int main()
{
    unsigned int n = 10, k = 2;
    if (bitAtGivenPosSetOrUnset(n, k))
        cout << "Set";
    else
        cout << "Unset";
    return 0;
}

                    

Java

// Java program to
// check the set bit
// at kth position
import java.io.*;
 
class GFG {
     
// function to check whether
// the bit at given position
// is set or unset
static int bitAtGivenPosSetOrUnset
                   ( int n, int k)
{
 
    // to shift the kth bit
    // at 1st position
    int New_num = 1 << (k - 1);
  
    // Since, last bit is now
    // kth bit, so doing AND with 1
    // will give result.
    return (New_num & n);
}
    public static void main (String[] args)
    {
         // K and n must be greater than 0
         int n = 10, k = 2;
          
    if (bitAtGivenPosSetOrUnset(n, k)==k)
        System.out.println("Set");
    else
        System.out.println("Unset");
    }
}
 
//This code is contributed by sam snehil

                    

Python3

# python implementation to check
# whether the bit at given
# position is set or unset
# by Using left shift operator
 
import math
#function to check whether the bit
# at given position is set or unset
# by Using left shift operator
def bitAtGivenPosSetOrUnset(  n,  k):
     New_num = 1 << (k - 1)
      
     # returning result
     return (New_num & n)
 
# Driver code
n = 10
k = 2
if (bitAtGivenPosSetOrUnset(n, k)):
     print("Set")
else:
    print("Unset")
 
#This code is contributed by sam snehil

                    

C#

// C# program to check the set bit
// at kth position Using left shift operator
using System;
 
class GFG {
    // Function using left shift operator
    static int bitAtGivenPosSetOrUnset(
                           int n, int k)
    {
     
        int New_num = 1 << (k - 1);
     
        // Returning result
        return (New_num & n);
    }
     
    // Driver code
    public static void Main ()
    {
         
        int n = 10, k = 2;
         
        if (bitAtGivenPosSetOrUnset(n, k) != 0)
            Console.Write("Set");
        else
            Console.Write("Unset");
    }
}
 
// This code is contributed by Sam snehil.

                    

Javascript

// javascript program to check set bit using left shift operator
 
// function using left shift operator
function bitAtGivenPosSetOrUnset
                   (n, k)
{
   
    let New_num = 1 << (k -1);
    
    // Returning result
    return (New_num & n);
}
 
// Driver Function
// K and n must be greater than 0
     let n = 10, k = 2;
            
    if (bitAtGivenPosSetOrUnset(n, k))
        console.log("Set");
    else
        console.log('Unset')
     
    // This code is contributed by sam snehil.

                    

PHP

<?php
// PHP implementation to check whether the bit
// at given position is set or unset Using left shift operator
 
// function demonstrate left shift operator working
function bitAtGivenPosSetOrUnset($n, $k)
{
    $New_num = 1 << ($k - 1);
 
    return ($New_num & $n);
}
 
    // Driver Code
    $n = 10;
    $k = 2;
    if (bitAtGivenPosSetOrUnset($n, $k))
        echo "Set";
    else
        echo "Unset";
         
// This code is contributed by Sam snehil
?>

                    

Output:

Set

Time Complexity: O(1)

Auxiliary Space: O(1)

Approach#3: We can use power of 2 to check if the Bitwise AND of ‘(k-1)th power of 2’ and num is 1 or not. If it is 1 , the bit is set else unset.

C++

// C++ implementation of the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if the kth bit is set
// or not in num
bool bitAtGivenPosSetOrUnset(int n, int k)
{
    int new_num = pow(2,(k -1));
    int x=new_num & n;//storing bitwise AND
   
    if(x > 0)
    { return true;// return true if kth bit is set
    }
    else{
    return false; // else return false
  }
}
// Drive Code
int main()
{
    int n = 10, k = 2;
   
    // Function call
    if (bitAtGivenPosSetOrUnset(n, k))
        cout << "Set";
    else
        cout << "Unset";
    return 0;
}
 
// This code is contributed by nikhilsainiofficial546

                    

Java

import java.lang.Math;
 
public class Main {
 
  // Function to check if the kth bit is set
  // or not in num
  static boolean bitAtGivenPosSetOrUnset(int n, int k)
  {
    int new_num = (int) Math.pow(2, k - 1);
    int x = new_num & n;  // storing bitwise AND
 
    if (x > 0) {
      return true// return true if kth bit is set
    }
    else {
      return false// else return false
    }
  }
 
  // Driver Code
  public static void main(String[] args) {
    int n = 10, k = 2;
 
    // Function call
    if (bitAtGivenPosSetOrUnset(n, k))
      System.out.println("Set");
    else
      System.out.println("Unset");
  }
}

                    

Python3

# Python3 implementation of the above approach
 
# Function to check if the kth bit is set or not in n
def bitAtGivenPosSetOrUnset(n, k):
     
    new_num = pow(2, k-1)
     
    # Check if the kth bit is set or not by performing bitwise AND
    x = new_num & n
     
    # Return True if kth bit is set, else False
    if x > 0:
        return True
    else:
        return False
       
 
# Drive Code
n = 10
k = 2
 
# Function call
if bitAtGivenPosSetOrUnset(n, k):
    print("Set")
else:
    print("Unset")
 
  # This code is contributed by nikhilsainiofficial546

                    

C#

using System;
 
public class Program
{
    // Function to check if the kth bit is set
// or not in num
public static bool bitAtGivenPosSetOrUnset(int n, int k)
{
    int new_num = (int)Math.Pow(2, (k - 1));
    int x = new_num & n;//storing bitwise AND
 
    if (x > 0)
    {
        return true;// return true if kth bit is set
    }
    else
    {
        return false; // else return false
    }
}
    // Drive Code
    public static void Main()
    {
        int n = 10, k = 2;
        // Function call
        if (bitAtGivenPosSetOrUnset(n, k))
            Console.WriteLine("Set");
        else
            Console.WriteLine("Unset");
    }
}
//This code is contributed by NarasingaNikhil

                    

Javascript

// Javascript implementation of the above approach
 
// Function to check if the kth bit is set or not in num
function bitAtGivenPosSetOrUnset(n, k) {
  const new_num = Math.pow(2, k - 1);
  const x = new_num & n; // storing bitwise AND
 
  if (x > 0) {
    return true; // return true if kth bit is set
  } else {
    return false; // else return false
  }
}
 
// Drive Code
const n = 10, k = 2;
 
// Function call
if (bitAtGivenPosSetOrUnset(n, k)) {
  console.log("Set");
} else {
  console.log("Unset");
}

                    

Output
Set

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



Last Updated : 07 Mar, 2023
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