Find value of k-th bit in binary representation

Given a number n and k (1 <= k <= 32), find the value of k-th bit in the binary representation of n. Bits are numbered from right (Least Significant Bit) to left (Most Significant Bit).

Examples :

Input :  
n = 13, k = 2
Output : 
0
Explanation:
Binary representation of 13 is 1101.
Second bit from right is 0.

Input :  
n = 14, k = 3
Output : 
1
Explanation:
Binary representation of 14 is 1110.
Third bit from right is 1.

Approach 1:
1) Find a number with all 0s except k-th position. We get this number using ( 1 << (k-1) ). For example if k = 3, then ( 1 << 2) gives us (00..00100). 
2) Do bitwise and of above-obtained number with n to find if k-th bit in n is set or not. 

Below is the implementation of the above approach:

C++

// CPP program to find k-th bit from right
#include <bits/stdc++.h>
using namespace std;

void printKthBit(unsigned int n, unsigned int k)
{
    cout << ((n & (1 << (k - 1))) >> (k - 1));
}

// Driver Code
int main()
{
    unsigned int n = 13, k = 2;

    // Function Call
    printKthBit(n, k);
    return 0;
}

Java

// Java program to find
// k-th bit from right
import java.io.*;

class GFG {
    static void printKthBit(long n, long k)
    {
        System.out.println(
            ((n & (1 << (k - 1))) >> (k - 1)));
    }

    // Driver Code
    public static void main(String[] args)
    {
        long n = 13, k = 2;

        // Function Call
        printKthBit(n, k);
    }
}

// This code is contributed by anuj_67.

Python3

# Python 3 program to find
# k-th bit from right

def printKthBit(n, k):

    print((n & (1 << (k - 1))) >> (k - 1))

# Driver Code
n = 13
k = 2

# Function Call
printKthBit(n, k)

# This code is contributed by Smitha

C#

// C# program to find k-th bit from right
using System;

class GFG {

    static void printKthBit(long n, long k)
    {
        Console.WriteLine((n & (1 << (k - 1))) >> (k - 1));
    }

    // Driver Code
    public static void Main()
    {
        long n = 13, k = 2;

        // Function Call
        printKthBit(n, k);
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// PHP program to find
// k-th bit from right

function printKthBit($n, $k)
{
    echo ($n & (1 << ($k - 1)));
}

// Driver Code
$n = 13; $k = 2;
printKthBit($n, $k);

// This code is contributed
// by anuj_67.
?>

Javascript

<script>

// JavaScript program to find k-th bit from right

function printKthBit(n, k)
{
    document.write((n & (1 << (k - 1))) >> (k - 1));
}

// Driver Code
var n = 13, k = 2;

// Function Call
printKthBit(n, k);

</script>

Output

0

Time Complexity: O(1), the code will run in a constant time.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Approach 2:

1) Right shift the number (k-1) times, so that the kth bit comes at the right (Least Significant Bit). We can do this by n >> (k-1). For example for number 37(100101) and k = 3, ( 32 >> (3-1) ) will be equal to 9(1001).

2) Then just check the last bit is set or not. We can easily do this by checking if number is odd or even.

Below is the implementation of the above approach:

C++

// CPP program to find k-th bit from right
#include <bits/stdc++.h>
using namespace std;

void printKthBit(unsigned int n, unsigned int k)
{
    cout << (n >> (k-1)) % 2;
}

// Driver Code
int main()
{
    unsigned int n = 37, k = 3;

    // Function Call
    printKthBit(n, k);
    return 0;
}
// This code is contributed by priyam26.

Java

// Java program to find k-th bit from right
import java.io.*;

class GFG {
    static void printKthBit(long n, long k)
    {
        System.out.println((n >> (k-1)) % 2);
    }

    // Driver Code
    public static void main(String[] args)
    {
        long n = 37, k = 3;

        // Function Call
        printKthBit(n, k);
    }
}

// This code is contributed by priyam26.

Python3

# Python 3 program to find k-th bit from right

def printKthBit(n, k):

    print((n >> (k-1)) % 2)

# Driver Code
n = 37
k = 3

# Function Call
printKthBit(n, k)

# This code is contributed by priyam26.

C#

// C# program to find k-th bit from right

using System;
using System.Collections.Generic;

class GFG {
    static void printKthBit(int n, int k)
    {
        Console.WriteLine((n >> (k-1)) % 2);
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int n = 37, k = 3;

        // Function Call
        printKthBit(n, k);
    }
}

// This code is contributed by phasing17.

Javascript

<script>

// JavaScript program to find k-th bit from right

function printKthBit(n, k)
{
    document.write((n >> (k-1)) % 2);
}

// Driver Code
var n = 37, k = 3;

// Function Call
printKthBit(n, k);

// This code is contributed by priyam26.
</script>

Output

1

Time Complexity: O(1), the code will run in a constant time.
Auxiliary Space: O(1), no extra space is required, so it is a constant.