Given a number x and two positions (from the right side) in the binary representation of x, write a function that swaps n bits at the given two positions and returns the result. It is also given that the two sets of bits do not overlap.
Method 1
Let p1 and p2 be the two given positions.
Example 1
Input: x = 47 (00101111) p1 = 1 (Start from the second bit from the right side) p2 = 5 (Start from the 6th bit from the right side) n = 3 (No of bits to be swapped) Output: 227 (11100011) The 3 bits starting from the second bit (from the right side) are swapped with 3 bits starting from 6th position (from the right side)
Example 2
Input: x = 28 (11100) p1 = 0 (Start from first bit from right side) p2 = 3 (Start from 4th bit from right side) n = 2 (No of bits to be swapped) Output: 7 (00111) The 2 bits starting from 0th position (from right side) are swapped with 2 bits starting from 4th position (from right side)
Solution
We need to swap two sets of bits. XOR can be used in a similar way as it is used to swap 2 numbers. Following is the algorithm.
1) Move all bits of the first set to the rightmost side set1 = (x >> p1) & ((1U << n) - 1) Here the expression (1U << n) - 1 gives a number that contains last n bits set and other bits as 0. We do & with this expression so that bits other than the last n bits become 0. 2) Move all bits of second set to rightmost side set2 = (x >> p2) & ((1U << n) - 1) 3) XOR the two sets of bits xor = (set1 ^ set2) 4) Put the xor bits back to their original positions. xor = (xor << p1) | (xor << p2) 5) Finally, XOR the xor with original number so that the two sets are swapped. result = x ^ xor
Implementation:
C++
// C++ Program to swap bits// in a given number#include <bits/stdc++.h>using namespace std;int swapBits(unsigned int x, unsigned int p1, unsigned int p2, unsigned int n){ /* Move all bits of first set to rightmost side */ unsigned int set1 = (x >> p1) & ((1U << n) - 1); /* Move all bits of second set to rightmost side */ unsigned int set2 = (x >> p2) & ((1U << n) - 1); /* Xor the two sets */ unsigned int Xor = (set1 ^ set2); /* Put the Xor bits back to their original positions */ Xor = (Xor << p1) | (Xor << p2); /* Xor the 'Xor' with the original number so that the two sets are swapped */ unsigned int result = x ^ Xor; return result;}/* Driver code*/int main(){ int res = swapBits(28, 0, 3, 2); cout << "Result = " << res; return 0;}// This code is contributed by rathbhupendra |
C
// C Program to swap bits// in a given number#include <stdio.h>int swapBits(unsigned int x, unsigned int p1, unsigned int p2, unsigned int n){ /* Move all bits of first set to rightmost side */ unsigned int set1 = (x >> p1) & ((1U << n) - 1); /* Move all bits of second set to rightmost side */ unsigned int set2 = (x >> p2) & ((1U << n) - 1); /* XOR the two sets */ unsigned int xor = (set1 ^ set2); /* Put the xor bits back to their original positions */ xor = (xor << p1) | (xor << p2); /* XOR the 'xor' with the original number so that the two sets are swapped */ unsigned int result = x ^ xor; return result;}/* Driver program to test above function*/int main(){ int res = swapBits(28, 0, 3, 2); printf("\nResult = %d ", res); return 0;} |
Java
// Java Program to swap bits// in a given numberclass GFG { static int swapBits(int x, int p1, int p2, int n) { // Move all bits of first set // to rightmost side int set1 = (x >> p1) & ((1 << n) - 1); // Move all bits of second set // to rightmost side int set2 = (x >> p2) & ((1 << n) - 1); // XOR the two sets int xor = (set1 ^ set2); // Put the xor bits back to // their original positions xor = (xor << p1) | (xor << p2); // XOR the 'xor' with the original number // so that the two sets are swapped int result = x ^ xor; return result; } // Driver program public static void main(String[] args) { int res = swapBits(28, 0, 3, 2); System.out.println("Result = " + res); }}// This code is contributed by prerna saini. |
Python3
# Python program to# swap bits in a given numberdef swapBits(x, p1, p2, n): # Move all bits of first # set to rightmost side set1 = (x >> p1) & ((1<< n) - 1) # Move all bits of second # set to rightmost side set2 = (x >> p2) & ((1 << n) - 1) # XOR the two sets xor = (set1 ^ set2) # Put the xor bits back # to their original positions xor = (xor << p1) | (xor << p2) # XOR the 'xor' with the # original number so that the # two sets are swapped result = x ^ xor return result # Driver coderes = swapBits(28, 0, 3, 2)print("Result =", res)# This code is contributed# by Anant Agarwal. |
C#
// C# Program to swap bits// in a given numberusing System;class GFG { static int swapBits(int x, int p1, int p2, int n) { // Move all bits of first // set to rightmost side int set1 = (x >> p1) & ((1 << n) - 1); // Move all bits of second set // set to rightmost side int set2 = (x >> p2) & ((1 << n) - 1); // XOR the two sets int xor = (set1 ^ set2); // Put the xor bits back to // their original positions xor = (xor << p1) | (xor << p2); // XOR the 'xor' with the original number // so that the two sets are swapped int result = x ^ xor; return result; } // Driver program public static void Main() { int res = swapBits(28, 0, 3, 2); Console.WriteLine("Result = " + res); }}// This code is contributed by vt_m. |
PHP
<?php// PHP Program to swap bits // in a given number// function returns// the swapped bitsfunction swapBits($x, $p1, $p2, $n){ // Move all bits of first // set to rightmost side $set1 = ($x >> $p1) & ((1 << $n) - 1); // Move all bits of second // set to rightmost side $set2 = ($x >> $p2) & ((1 << $n) - 1); // XOR the two sets $xor = ($set1 ^ $set2); // Put the xor bits back to // their original positions $xor = ($xor << $p1) | ($xor << $p2); // XOR the 'xor' with the // original number so that // the two sets are swapped $result = $x ^ $xor; return $result;} // Driver Code $res = swapBits(28, 0, 3, 2); echo "\nResult = ", $res; // This code is contributed by anuj_67.?> |
Javascript
<script>// Javascript Program to swap bits// in a given number function swapBits(x, p1, p2, n) { // Move all bits of first set // to rightmost side let set1 = (x >> p1) & ((1 << n) - 1); // Move all bits of second set // to rightmost side let set2 = (x >> p2) & ((1 << n) - 1); // XOR the two sets let xor = (set1 ^ set2); // Put the xor bits back to // their original positions xor = (xor << p1) | (xor << p2); // XOR the 'xor' with the original number // so that the two sets are swapped let result = x ^ xor; return result; }// Driver Code let res = swapBits(28, 0, 3, 2); document.write("Result = " + res);</script> |
Output
Result = 7
Time Complexity: O(1), as we are using constant-time operations.
Auxiliary Space: O(1), as we are not using any extra space.
Following is a shorter implementation of the same logic
C++
// C++ program of the above approach#include <bits/stdc++.h>using namespace std;int swapBits(int x, int p1, int p2, int n){ /* xor contains xor of two sets */ int xor = ((x >> p1) ^ (x >> p2)) & ((1U << n) - 1); /* To swap two sets, we need to again XOR the xor with * original sets */ return x ^ ((xor << p1) | (xor << p2));}// This code is contributed by splevel62. |
C
int swapBits(unsigned int x, unsigned int p1, unsigned int p2, unsigned int n){ /* xor contains xor of two sets */ unsigned int xor = ((x >> p1) ^ (x >> p2)) & ((1U << n) - 1); /* To swap two sets, we need to again XOR the xor with original sets */ return x ^ ( (xor << p1) | (xor << p2));} |
Java
static int swapBits(int x, int p1, int p2, int n){ /* xor contains xor of two sets */ int xor = ((x >> p1) ^ (x >> p2)) & ((1U << n) - 1); /* To swap two sets, we need to again XOR the xor with * original sets */ return x ^ ((xor << p1) | (xor << p2));}// This code is contributed by subhammahato348. |
Python3
# Python code to implement the approachdef swapBits(x, p1, p2, n) : # xor contains xor of two sets xor = (((x >> p1) ^ (x >> p2)) & ((1 << n) - 1)) # To swap two sets, we need to again XOR the xor with original sets return x ^ ( (xor << p1) | (xor << p2)) # This code is contributed by sanjoy_62. |
C#
static int swapBits(int x, int p1, int p2, int n){ /* xor contains xor of two sets */ int xor = ((x >> p1) ^ (x >> p2)) & ((1U << n) - 1); /* To swap two sets, we need to again XOR the xor with * original sets */ return x ^ ((xor << p1) | (xor << p2));}// This code is contributed by subhammahato348. |
Javascript
<script> // JavaScript code for the above approachfunction swapBits(x, p1, p2, n){ /* xor contains xor of two sets */ let xor = ((x >> p1) ^ (x >> p2)) & ((1 << n) - 1); /* To swap two sets, we need to again XOR the xor with * original sets */ return x ^ ((xor << p1) | (xor << p2));}// This code is contributed by avijitmondal1998.</script> |
Time Complexity: O(1), as we are using constant-time operations.
Auxiliary Space: O(1), as we are not using any extra space.
Method 2 –
This solution focuses on calculating the values of bits to be swapped using AND gate. Then we can set/unset those bits based on whether the bits are to be swapped. For the number of bits to be swapped (n) –
- Calculate shift1 = The value after setting bit at p1 position to 1
- Calculate shift2 = The value after setting bit at p2 position to 1
- value1 = Number to check if num at position p1 is set or not.
- value2 = Number to check if num at position p2 is set or not.
- If value1 and value2 are different is when we have to swap the bits.
Example:
[28 0 3 2] num=28 (11100) p1=0 p2=3 n=2 Given = 11100 Required output = 00111 i.e. (00)1(11) msb 2 bits replaced with lsb 2 bits n=2 p1=0, p2=3 shift1= 1, shift2= 1000 value1= 0, value2= 1000 After swap num= 10101 n=3 p1=1, p2=4 shift1= 10, shift2= 10000 value1= 0, value2= 10000 After swap num= 00111
Implementation
C++
#include <iostream>using namespace std;int swapBits(unsigned int num, unsigned int p1, unsigned int p2, unsigned int n){ int shift1, shift2, value1, value2; while (n--) { // Setting bit at p1 position to 1 shift1 = 1 << p1; // Setting bit at p2 position to 1 shift2 = 1 << p2; // value1 and value2 will have 0 if num // at the respective positions - p1 and p2 is 0. value1 = ((num & shift1)); value2 = ((num & shift2)); // check if value1 and value2 are different // i.e. at one position bit is set and other it is not if ((!value1 && value2) || (!value2 && value1)) { // if bit at p1 position is set if (value1) { // unset bit at p1 position num = num & (~shift1); // set bit at p2 position num = num | shift2; } // if bit at p2 position is set else { // set bit at p2 position num = num & (~shift2); // unset bit at p2 position num = num | shift1; } } p1++; p2++; } // return final result return num;}/* Driver code*/int main(){ int res = swapBits(28, 0, 3, 2); cout << "Result = " << res; return 0;} |
Java
class GFG{ static int swapBits(int num, int p1, int p2, int n) { int shift1, shift2, value1, value2; while (n-- > 0) { // Setting bit at p1 position to 1 shift1 = 1 << p1; // Setting bit at p2 position to 1 shift2 = 1 << p2; // value1 and value2 will have 0 if num // at the respective positions - p1 and p2 is 0. value1 = ((num & shift1)); value2 = ((num & shift2)); // check if value1 and value2 are different // i.e. at one position bit is set and other it is not if ((value1 == 0 && value2 != 0) || (value2 == 0 && value1 != 0)) { // if bit at p1 position is set if (value1 != 0) { // unset bit at p1 position num = num & (~shift1); // set bit at p2 position num = num | shift2; } // if bit at p2 position is set else { // set bit at p2 position num = num & (~shift2); // unset bit at p2 position num = num | shift1; } } p1++; p2++; } // return final result return num; } // Driver code public static void main(String[] args) { int res = swapBits(28, 0, 3, 2); System.out.println("Result = " + res); }}// This code is contributed by divyeshrabadiya07 |
Python3
def swapBits(num, p1, p2, n): shift1 = 0 shift2 = 0 value1 = 0 value2 = 0 while(n > 0): # Setting bit at p1 position to 1 shift1 = 1 << p1 # Setting bit at p2 position to 1 shift2 = 1 << p2 # value1 and value2 will have 0 if num # at the respective positions - p1 and p2 is 0. value1 = ((num & shift1)) value2 = ((num & shift2)) # check if value1 and value2 are different # i.e. at one position bit is set and other it is not if((value1 == 0 and value2 != 0) or (value2 == 0 and value1 != 0)): # if bit at p1 position is set if(value1 != 0): # unset bit at p1 position num = num & (~shift1) # set bit at p2 position num = num | shift2 # if bit at p2 position is set else: # set bit at p2 position num = num & (~shift2) # unset bit at p2 position num = num | shift1 p1 += 1 p2 += 1 n -= 1 # return final result return num# Driver coderes = swapBits(28, 0, 3, 2)print("Result =", res)# This code is contributed by avanitrachhadiya2155 |
C#
using System;class GFG { static int swapBits(int num, int p1, int p2, int n) { int shift1, shift2, value1, value2; while (n-- > 0) { // Setting bit at p1 position to 1 shift1 = 1 << p1; // Setting bit at p2 position to 1 shift2 = 1 << p2; // value1 and value2 will have 0 if num // at the respective positions - p1 and p2 is 0. value1 = ((num & shift1)); value2 = ((num & shift2)); // check if value1 and value2 are different // i.e. at one position bit is set and other it is not if ((value1 == 0 && value2 != 0) || (value2 == 0 && value1 != 0)) { // if bit at p1 position is set if (value1 != 0) { // unset bit at p1 position num = num & (~shift1); // set bit at p2 position num = num | shift2; } // if bit at p2 position is set else { // set bit at p2 position num = num & (~shift2); // unset bit at p2 position num = num | shift1; } } p1++; p2++; } // return final result return num; } // Driver code static void Main() { int res = swapBits(28, 0, 3, 2); Console.WriteLine("Result = " + res); }}// This code is contributed by divyesh072019 |
Javascript
<script>function swapBits(num, p1, p2, n){ let shift1, shift2, value1, value2; while (n-- > 0) { // Setting bit at p1 position to 1 shift1 = 1 << p1; // Setting bit at p2 position to 1 shift2 = 1 << p2; // value1 and value2 will have 0 if num // at the respective positions - p1 and p2 is 0. value1 = ((num & shift1)); value2 = ((num & shift2)); // Check if value1 and value2 are different // i.e. at one position bit is set and // other it is not if ((value1 == 0 && value2 != 0) || (value2 == 0 && value1 != 0)) { // If bit at p1 position is set if (value1 != 0) { // Unset bit at p1 position num = num & (~shift1); // Set bit at p2 position num = num | shift2; } // If bit at p2 position is set else { // Set bit at p2 position num = num & (~shift2); // Unset bit at p2 position num = num | shift1; } } p1++; p2++; } // Return final result return num;}// Driver codelet res = swapBits(28, 0, 3, 2);document.write("Result = " + res);// This code is contributed by suresh07 </script> |
Output
Result = 7
Time Complexity: O(N), as we are using a loop to traverse N times. Where N is the number of bits to be swapped.
Auxiliary Space: O(1), as we are not using any extra space.
References:
Swapping individual bits with XOR
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