Minimum bit swaps between given numbers to make their Bitwise OR equal to Bitwise AND
Given two positive integers A and B, the task is to calculate the minimum number of operations required such that Bitwise OR of A and B equals Bitwise AND of A and B are equal i.e (A&B)=(A|B), where, in each operation two indices i and j are chosen and the ith bit of A is swapped with the jth bit of B. If it is not possible to make (A&B)=(A|B), print -1.
Examples:
Input: A = 1, B = 2
Output: 2
Explanation:
A10 ≡ 012, B10 ≡ 102
The following sequence of moves can be performed:
- i = 1, j = 1⇒ A = 11, B = 00 (A|B = 3, A&B = 0).
- i = 1, j = 0⇒ A = 01, B = 01 (A|B = 1, A&B = 1).
Thus, 2 moves are required.
Input: A = 27, B = 5
Output: 3
Explanation:
A10 ≡ 110112, B10 ≡ 001012
The following sequence of moves can be performed:
- i = 4, j = 3⇒ A = 01011, B = 01101 (A|B = 15, A&B = 9).
- i = 2, j = 2⇒ A = 01111, B = 01001 (A|B = 15, A&B = 9).
- i = 2, j = 1⇒ A = 01011, B = 01011 (A|B = 11, A&B = 11).
Thus, 3 moves are required.
Approach:
Observation: The main observation to solve this problem is that for (A&B)=(A|B) is that A must be equal to B because if only two bits are set, then only their Bitwise AND and Bitwise OR are equal.
Follow the below steps to solve the problem:
- Count the number of total set bits in A and B.
- If the count is odd, the two numbers cannot be made equal, so print -1.
- Initialize two counters oneZero=0 and zeroOne=0
- Traverse through the bits of A and B, and do the following:
- If current bit of A is set and current bit of B is unset i.e (1, 0), increment oneZero.
- If current bit of A is unset and current bit of B is set i.e (0, 1), increment zeroOne.
- To minimize the number of operations required, it is optimal to choose two (1, 0) or two (0, 1) indices and swap either one of them, i.e only half of oneZero and zeroOne operations are required.
- If oneZero is odd(which means zeroOne is also odd), two more operations would be required to turn (0, 1) and a (1, 0) to (1, 1) and (0, 0)
- So, the final answer is (oneZero/2)+(zeroOne/2)+(oneZero%2?2:0).
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int countSetBits( int n)
{
int count = 0;
while (n) {
n = n & (n - 1);
count++;
}
return count;
}
int minOperations( int A, int B)
{
int cnt1 = 0, cnt2 = 0;
cnt1 += countSetBits(A);
cnt2 += countSetBits(B);
if ((cnt1 + cnt2) % 2 != 0)
return -1;
int oneZero = 0, zeroOne = 0;
int ans = 0;
for ( int i = 0; i < max(cnt1, cnt2); i++) {
int bitpos = 1 << i;
if ((!(bitpos & A)) && (bitpos & B))
zeroOne++;
if ((bitpos & A) && (!(bitpos & B)))
oneZero++;
}
ans = (zeroOne / 2) + (oneZero / 2);
if (zeroOne % 2 != 0)
ans += 2;
return ans;
}
int main()
{
int A = 27, B = 5;
cout << minOperations(A, B);
return 0;
}
|
Java
import java.io.*;
class GFG{
static int countSetBits( int n)
{
int count = 0 ;
while (n != 0 ) {
n = n & (n - 1 );
count++;
}
return count;
}
static int minOperations( int A, int B)
{
int cnt1 = 0 , cnt2 = 0 ;
cnt1 += countSetBits(A);
cnt2 += countSetBits(B);
if ((cnt1 + cnt2) % 2 != 0 )
return - 1 ;
int oneZero = 0 , zeroOne = 0 ;
int ans = 0 ;
for ( int i = 0 ; i < Math.max(cnt1, cnt2); i++) {
int bitpos = 1 << i;
if (((bitpos & A) == 0 ) && ((bitpos & B) != 0 ))
zeroOne++;
if (((bitpos & A) != 0 ) && ((bitpos & B) == 0 ))
oneZero++;
}
ans = (zeroOne / 2 ) + (oneZero / 2 );
if (zeroOne % 2 != 0 )
ans += 2 ;
return ans;
}
public static void main(String args[])
{
int A = 27 , B = 5 ;
System.out.println( minOperations(A, B));
}
}
|
Python3
def countSetBits(n):
count = 0
while (n):
n = n & (n - 1 )
count + = 1
return count
def minOperations(A, B):
cnt1 = 0
cnt2 = 0
cnt1 + = countSetBits(A)
cnt2 + = countSetBits(B)
if ((cnt1 + cnt2) % 2 ! = 0 ):
return - 1
oneZero = 0
zeroOne = 0
ans = 0
for i in range ( max (cnt1, cnt2)):
bitpos = 1 << i
if (( not (bitpos & A)) and (bitpos & B)):
zeroOne + = 1
if ((bitpos & A) and ( not (bitpos & B))):
oneZero + = 1
ans = (zeroOne / / 2 ) + (oneZero / / 2 )
if (zeroOne % 2 ! = 0 ):
ans + = 2
return ans
if __name__ = = '__main__' :
A = 27
B = 5
print (minOperations(A, B))
|
C#
using System;
using System.Collections.Generic;
class GFG{
static int countSetBits( int n)
{
int count = 0;
while (n > 0)
{
n = n & (n - 1);
count++;
}
return count;
}
static int minOperations( int A, int B)
{
int cnt1 = 0, cnt2 = 0;
cnt1 += countSetBits(A);
cnt2 += countSetBits(B);
if ((cnt1 + cnt2) % 2 != 0)
return -1;
int oneZero = 0, zeroOne = 0;
int ans = 0;
for ( int i = 0; i < Math.Max(cnt1, cnt2); i++)
{
int bitpos = 1 << i;
if (((bitpos & A) == 0) && (bitpos & B) != 0)
zeroOne++;
if ((bitpos & A) != 0 && ((bitpos & B) == 0))
oneZero++;
}
ans = (zeroOne / 2) + (oneZero / 2);
if (zeroOne % 2 != 0)
ans += 2;
return ans;
}
public static void Main()
{
int A = 27, B = 5;
Console.Write(minOperations(A, B));
}
}
|
Javascript
<script>
function countSetBits(n)
{
let count = 0;
while (n) {
n = n & (n - 1);
count++;
}
return count;
}
function minOperations(A, B)
{
let cnt1 = 0, cnt2 = 0;
cnt1 += countSetBits(A);
cnt2 += countSetBits(B);
if ((cnt1 + cnt2) % 2 != 0)
return -1;
let oneZero = 0, zeroOne = 0;
let ans = 0;
for (let i = 0; i < Math.max(cnt1, cnt2); i++) {
let bitpos = 1 << i;
if ((!(bitpos & A)) && (bitpos & B))
zeroOne++;
if ((bitpos & A) && (!(bitpos & B)))
oneZero++;
}
ans = parseInt(zeroOne / 2) + parseInt(oneZero / 2);
if (zeroOne % 2 != 0)
ans += 2;
return ans;
}
let A = 27, B = 5;
document.write(minOperations(A, B));
</script>
|
Time Complexity: O(Log2N)
Auxiliary space: O(1)
Last Updated :
13 Jul, 2021
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...