Check whether the bit at given position is set or unset

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
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
int main()
{
    unsigned int n = 10, k = 2;
    if (bitAtGivenPosSetOrUnset(n, k))
        cout << "Set";
    else
        cout << "Unset";
    return 0;
}

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.

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.

PHP

<?php
// PHP implementation to check whether the bit
// at given position is set or unset Using left shift operator

// function demonstrate left shif 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
?>

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

Output:

Set

Time Complexity: O(1)

Auxiliary Space: O(1)