Open In App

How to swap two bits in a given integer?

Improve
Improve
Like Article
Like
Save
Share
Report

Given an integer n and two-bit positions p1 and p2 inside it, swap bits at the given positions. The given positions are from the least significant bit (lsb). For example, the position for lsb is 0.
Examples: 

Input: n = 28, p1 = 0, p2 = 3
Output: 21
Explanation: 28 in binary is 11100.  If we swap 0’th and 3rd digits, we get 10101 which is 21 in decimal.

Input: n = 20, p1 = 2, p2 = 3
Output: 24

We strongly recommend you minimize your browser and try this yourself first.
Method 1: 
The idea is to first find the bits, then use XOR based swapping concept, i..e., to swap two numbers ‘x’ and ‘y’, we do x = x ^ y,  y = y ^ x, and x = x ^ y.

Below is the implementation of the above idea

C++




// C++ program to swap bits in an integer
#include<bits/stdc++.h>
using namespace std;
 
// This function swaps bit at positions p1 and p2 in an integer n
int swapBits(unsigned int n, unsigned int p1, unsigned int p2)
{
    /* Move p1'th to rightmost side */
    unsigned int bit1 =  (n >> p1) & 1;
 
    /* Move p2'th to rightmost side */
    unsigned int bit2 =  (n >> p2) & 1;
 
    /* XOR the two bits */
    unsigned int x = (bit1 ^ bit2);
 
    /* Put the xor bit back to their original positions */
    x = (x << p1) | (x << p2);
 
    /* XOR 'x' with the original number so that the
       two sets are swapped */
    unsigned int result = n ^ x;
}
 
/* Driver program to test above function*/
int main()
{
    int res =  swapBits(28, 0, 3);
    cout<<"Result = "<< res<<" ";
    return 0;
}
 
// This code is contributed by pratham76.


C




// C program to swap bits in an integer
#include<stdio.h>
 
// This function swaps bit at positions p1 and p2 in an integer n
int swapBits(unsigned int n, unsigned int p1, unsigned int p2)
{
    /* Move p1'th to rightmost side */
    unsigned int bit1 =  (n >> p1) & 1;
 
    /* Move p2'th to rightmost side */
    unsigned int bit2 =  (n >> p2) & 1;
 
    /* XOR the two bits */
    unsigned int x = (bit1 ^ bit2);
 
    /* Put the xor bit back to their original positions */
    x = (x << p1) | (x << p2);
 
    /* XOR 'x' with the original number so that the
       two sets are swapped */
    unsigned int result = n ^ x;
}
 
/* Driver program to test above function*/
int main()
{
    int res =  swapBits(28, 0, 3);
    printf("Result = %d ", res);
    return 0;
}


Java




// Java program to swap bits in an integer
import java.io.*;
 
class GFG
{
     
// This function swaps bit at
// positions p1 and p2 in an integer n
static int swapBits( int n, int p1, int p2)
{
    /* Move p1'th to rightmost side */
    int bit1 = (n >> p1) & 1;
 
    /* Move p2'th to rightmost side */
    int bit2 = (n >> p2) & 1;
 
    /* XOR the two bits */
    int x = (bit1 ^ bit2);
 
    /* Put the xor bit back to
    their original positions */
    x = (x << p1) | (x << p2);
 
    /* XOR 'x' with the original
    number so that the
    two sets are swapped */
    int result = n ^ x;
    return result;
}
 
    /* Driver code*/
    public static void main (String[] args)
    {
        int res = swapBits(28, 0, 3);
        System.out.println ("Result = " + res);
    }
}
 
// This code is contributed by ajit..


C#




// C# program to swap bits in an integer
using System;
class GFG
{
 
  // This function swaps bit at
  // positions p1 and p2 in an integer n
  static int swapBits( int n, int p1, int p2)
  {
 
    /* Move p1'th to rightmost side */
    int bit1 = (n >> p1) & 1;
 
    /* Move p2'th to rightmost side */
    int bit2 = (n >> p2) & 1;
 
    /* XOR the two bits */
    int x = (bit1 ^ bit2);
 
    /* Put the xor bit back to
    their original positions */
    x = (x << p1) | (x << p2);
 
    /* XOR 'x' with the original
    number so that the
    two sets are swapped */
    int result = n ^ x;
    return result;
  }
 
  /* Driver code*/
  public static void Main(string[] args)
  {
    int res = swapBits(28, 0, 3);
    Console.Write("Result = " + res);
  }
}
 
// This code is contributed by rutvik_56.


Javascript




<script>
// javascript program to swap bits in an integer
 
     
// This function swaps bit at
// positions p1 and p2 in an integer n
function swapBits(n , p1 , p2)
{
    /* Move p1'th to rightmost side */
    var bit1 = (n >> p1) & 1;
 
    /* Move p2'th to rightmost side */
    var bit2 = (n >> p2) & 1;
 
    /* XOR the two bits */
    var x = (bit1 ^ bit2);
 
    /* Put the xor bit back to
    their original positions */
    x = (x << p1) | (x << p2);
 
    /* XOR 'x' with the original
    number so that the
    two sets are swapped */
    var result = n ^ x;
    return result;
}
 
/* Driver code*/
var res = swapBits(28, 0, 3);
document.write("Result = " + res);
 
 
// This code contributed by Princi Singh
</script>


Python3




# Python3 program for the above approach
 
# This function swaps bit at positions p1 and p2 in an integer n
def swapBits(n, p1, p2):
 
    # Move p1'th to rightmost side
    bit1 = (n >> p1) & 1
 
    # Move p2'th to rightmost side
    bit2 = (n >> p2) & 1
 
    # XOR the two bits
    x = (bit1 ^ bit2)
 
    # Put the xor bit back to their original positions
    x = (x << p1) | (x << p2)
 
    # XOR 'x' with the original number so that the
    # two sets are swapped
    result = n ^ x
    return result
 
 
# Driver program to test above function
if __name__ == '__main__':
    res = swapBits(28, 0, 3)
    print("Result = ", res)
 
# This code is contributed by nirajgusain5


Output

Result = 21 

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

C++




//C++ code for swapping given bits of a number
#include<iostream>
using namespace std;
int swapBits(int n, int p1, int p2)
{
   //left-shift 1 p1 and p2 times
   //and using XOR
  if (((n & (1 << p1)) >> p1) ^ ((n & (1 << p2)) >> p2))
  {
    n ^= 1 << p1;
    n ^= 1 << p2;
  }
  return n;
}
 
//Driver Code
int main()
{
    cout << "Result = " << swapBits(28, 0, 3);
    return 0;
}


C




//C code for swapping given bits of a number
#include<stdio.h>
int swapBits(int n, int p1, int p2)
{
    //left-shift 1 p1 and p2 times
    //and using XOR
    if (((n & (1 << p1)) >> p1) ^ ((n & (1 << p2)) >> p2))
    {
      n ^= 1 << p1;
      n ^= 1 << p2;
    }
    return n;
}
 
//Driver Code
int main()
{
    printf("Result = %d", swapBits(28, 0, 3));
    return 0;
}


Java




// Java code for swapping
// given bits of a number
import java.util.*;
class Main{
   
public static int swapBits(int n,
                           int p1,
                           int p2)
{
  //left-shift 1 p1 and
  // p2 times and using XOR
  int temp = ((n & (1 << p1)) >> p1) ^ ((n & (1 << p2)) >> p2);
  if (temp >= 1)
  {
    n ^= 1 << p1;
    n ^= 1 << p2;
  }
  return n;
}   
 
// Driver code
public static void main(String[] args)
{
  System.out.print("Result = " +
                   swapBits(28, 0, 3));
}
}
 
// This code is contributed by divyeshrabadiya07


Python




# Python code for swapping given bits of a number
def swapBits(n, p1, p2):
 
  # left-shift 1 p1 and p2 times
  # and using XOR
    if ((n & (1 << p1)) >> p1) ^ ((n & (1 << p2)) >> p2):
        n ^= 1 << p1
        n ^= 1 << p2
    return n
 
# Driver Code
print("Result =",swapBits(28, 0, 3))
 
# This code is contributed by rag2127


C#




// C# code for swapping given bits of a number
using System;
class GFG {
 
    static int swapBits(int n, int p1, int p2)
    {
        // left-shift 1 p1 and p2 times
        // and using XOR
    int temp = ((n & (1 << p1)) >> p1) ^ ((n & (1 << p2)));
    if (temp >= 1)
    {
            n ^= 1 << p1;
            n ^= 1 << p2;
    }
    return n;
    }
 
    // Driver code
    static void Main()
    {
        Console.WriteLine("Result = " + swapBits(28, 0, 3));
    }
}
 
// This code is contributed by divyesh072019


Javascript




<script>
 
    // JavaScript code for swapping given bits of a number
     
    function swapBits(n, p1, p2)
    {
        temp = ((n & (1 << p1)) >> p1) ^ ((n & (1 << p2)) >> p2);
        if (temp >= 1)
        {
          n ^= 1 << p1;
          n ^= 1 << p2;
        }
        return n;
    }
 
    document.write("Result = " + swapBits(28, 0, 3));
     
</script>


Output

Result = 21

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



Last Updated : 22 Jul, 2022
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads