Count ways to generate pairs having Bitwise XOR and Bitwise AND equal to X and Y respectively
Given two integers X and Y, the task is to find the total number of ways to generate a pair of integers A and B such that Bitwise XOR and Bitwise AND between A and B is X and Y respectively
Examples:
Input: X = 2, Y = 5
Output: 2
Explanation:
The two possible pairs are (5, 7) and (7, 5).
Pair 1: (5, 7)
Bitwise AND = 5 & 7 = 2
Bitwise XOR = 5 ^ 7 = 5
Pair 2: (7, 5)
Bitwise AND = 7 & 5 = 2
Bitwise XOR = 7 ^ 5 = 5
Input: X = 7, Y = 5
Output: 0
Naive Approach: The simplest approach to solve the problem is to choose the maximum among X and Y and set all its bits and then check all possible pairs from 0 to that maximum number, say M. If for any pair of A and B, A & B and A?B becomes equal to X and Y respectively, then increment the count. Print the final value of count after checking for all possible pairs.
Time Complexity: O(M2), where M = max(X,Y)
Auxiliary Space: O(1)
Efficient Approach: The idea is to generate all possible combinations of bits at each position. There are 4 possibilities for the ith bit in X and Y which are as follows:
- If Xi = 0 and Yi = 1, then Ai = 1 and Bi = 1.
- If Xi = 1 and Yi = 1, then no possible bit assignment exist.
- If Xi = 0 and Yi = 0 then Ai = 0 and Bi = 0.
- If Xi = 1 and Yi = 0 then Ai = 0 and Bi = 1 or Ai = 1 and Bi = 0, where Ai, Bi, Xi, and Yi represent ith bit in each of them.
Follow the steps below to solve the problem:
- Initialize the counter as 1.
- For the ith bit, if Xi and Yi are equal to 1, then print 0.
- If at ith bit, Xi is 1 and Yi is 0 then multiply the counter by 2 as there are 2 options.
- After that, divide X and Y each time by 2.
- Repeat the above steps for every ith bit until both of them become 0 and print the value of the counter.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
int countOfPairs( int x, int y)
{
int counter = 1;
while (x || y) {
int bit1 = x % 2;
int bit2 = y % 2;
x >>= 1;
y >>= 1;
if (bit1 == 1 and bit2 == 0) {
counter *= 2;
continue ;
}
if (bit1 & bit2) {
counter = 0;
break ;
}
}
return counter;
}
int main()
{
int X = 2, Y = 5;
cout << countOfPairs(X, Y);
return 0;
}
|
Java
import java.util.*;
class GFG{
static int countOfPairs( int x, int y)
{
int counter = 1 ;
while (x > 0 || y > 0 )
{
int bit1 = x % 2 ;
int bit2 = y % 2 ;
x >>= 1 ;
y >>= 1 ;
if (bit1 == 1 && bit2 == 0 )
{
counter *= 2 ;
continue ;
}
if ((bit1 & bit2) > 0 )
{
counter = 0 ;
break ;
}
}
return counter;
}
public static void main(String[] args)
{
int X = 2 , Y = 5 ;
System.out.print(countOfPairs(X, Y));
}
}
|
Python3
def countOfPairs(x, y):
counter = 1
while (x or y):
bit1 = x % 2
bit2 = y % 2
x >> = 1
y >> = 1
if (bit1 = = 1 and bit2 = = 0 ):
counter * = 2
continue
if (bit1 & bit2):
counter = 0
break
return counter
X = 2
Y = 5
print (countOfPairs(X, Y))
|
C#
using System;
class GFG{
static int countOfPairs( int x, int y)
{
int counter = 1;
while (x > 0 || y > 0)
{
int bit1 = x % 2;
int bit2 = y % 2;
x >>= 1;
y >>= 1;
if (bit1 == 1 && bit2 == 0)
{
counter *= 2;
continue ;
}
if ((bit1 & bit2) > 0)
{
counter = 0;
break ;
}
}
return counter;
}
public static void Main(String[] args)
{
int X = 2, Y = 5;
Console.Write(countOfPairs(X, Y));
}
}
|
Javascript
<script>
function countOfPairs(x, y)
{
let counter = 1;
while (x > 0 || y > 0)
{
let bit1 = x % 2;
let bit2 = y % 2;
x >>= 1;
y >>= 1;
if (bit1 == 1 && bit2 == 0)
{
counter *= 2;
continue ;
}
if ((bit1 & bit2) > 0)
{
counter = 0;
break ;
}
}
return counter;
}
let X = 2, Y = 5;
document.write(countOfPairs(X, Y));
</script>
|
Time Complexity: O(log M), where M = max(X,Y), as we are using a loop to traverse and each time we are decrementing by floor division of 2.
Auxiliary Space: O(1), as we are not using any extra space.
Last Updated :
25 Apr, 2023
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