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++ 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 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 Program to swap bits // in a given number 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
// 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. |
# Python program to # swap bits in a given number def 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 )
print ( "Result =" , res)
# This code is contributed # by Anant Agarwal. |
// C# Program to swap bits // in a given number using 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 Program to swap bits // in a given number // function returns // the swapped bits function 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. ?> |
<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> |
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++ 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. |
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));
} |
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. |
# Python code to implement the approach def 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.
|
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. |
<script> // JavaScript code for the above approach
function 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
#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;
} |
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 |
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 code res = swapBits( 28 , 0 , 3 , 2 )
print ( "Result =" , res)
# This code is contributed by avanitrachhadiya2155 |
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 |
<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 code let res = swapBits(28, 0, 3, 2); document.write( "Result = " + res);
// This code is contributed by suresh07 </script> |
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