Check if a pair of integers from two ranges exists such that their Bitwise XOR exceeds both the ranges

Given two integers A and B, the task is to check if there exists two integers P and Q over the range [1, A] and [1, B] respectively such that Bitwise XOR of P and Q is greater than A and B. If found to be true, then print "Yes". Otherwise, print "No".

Examples:

Input: X = 2, Y = 2 
Output: Yes
Explanation:
By choosing the value of P and Q as 1 and 2 respectively, gives the Bitwise XOR of P and Q as 1^2 = 3 which is greater than Bitwise XOR of A and B A ^ B = 0.
Therefore, print Yes.

Input: X = 2, Y = 4
Output: No

Naive Approach: The simplest approach to solve the given problem is to generate all possible pairs of (P, Q) by traversing all integers from 1 to X and 1 to Y and check if there exists a pair such that their Bitwise XOR is greater than Bitwise XOR of X and Y, then print "Yes". Otherwise, print "No".

Below is the implementation of the above approach:

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if there exists any pair (P, Q) whose
// Bitwise XOR is greater than the Bitwise XOR of X and Y
void findWinner(int X, int Y)
{
    // Stores the Bitwise XOR of X & Y
    int playerA = (X ^ Y);
    bool flag = false;
    // Traverse all possible pairs
    for (int i = 1; i <= X; i++) {
        for (int j = 1; j <= Y; j++) {
            int val = (i ^ j);
            // If a pair exists
            if (val > playerA) {
                flag = true;
                break;
            }
        }
        if (flag)
            break;
    }
    // If a pair is found
    if (flag)
        cout << "Yes";
    else
        cout << "No";
}

// Driver Code
int main()
{
    int A = 2, B = 4;
    findWinner(A, B);
    return 0;
}

// C program for the above approach
#include <stdio.h>
#include <stdbool.h>

// Function to check if there exists any pair (P, Q) whose
// Bitwise XOR is greater than the Bitwise XOR of X and Y
void findWinner(int X, int Y)
{
    // Stores the Bitwise XOR of X & Y
    int playerA = (X ^ Y);
    bool flag = false;
    // Traverse all possible pairs
    for (int i = 1; i <= X; i++) {
        for (int j = 1; j <= Y; j++) {
            int val = (i ^ j);
            // If a pair exists
            if (val > playerA) {
                flag = true;
                break;
            }
        }
        if (flag)
            break;
    }
    // If a pair is found
    if (flag)
        printf("Yes");
    else
        printf("No");
}

// Driver Code
int main()
{
    int A = 2, B = 4;
    findWinner(A, B);
    return 0;
}

// This code is contributed by Sania Kumari Gupta

// Java program for the above approach
import java.lang.*;
import java.util.*;

class GFG{
    
// Function to check if there exists
// any pair (P, Q) whose Bitwise XOR
// is greater than the Bitwise XOR
// of X and Y
static void findWinner(int X, int Y)
{
    
    // Stores the Bitwise XOR of X & Y
    int playerA = (X ^ Y);

    boolean flag = false;

    // Traverse all possible pairs
    for(int i = 1; i <= X; i++) 
    {
        for(int j = 1; j <= Y; j++)
        {
            int val = (i ^ j);

            // If a pair exists
            if (val > playerA) 
            {
                flag = true;
                break;
            }
        }

        if (flag)
        {
            break;
        }
    }

    // If a pair is found
    if (flag) 
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}

// Driver code
public static void main(String[] args)
{ 
    int A = 2, B = 4;
    
    findWinner(A, B);
}
}

// This code is contributed by offbeat

# Python3 program for the above approach

# Function to check if there exists
# any pair (P, Q) whose Bitwise XOR
# is greater than the Bitwise XOR
# of X and Y
def findWinner(X, Y):
    
    # Stores the Bitwise XOR of X & Y
    playerA = (X ^ Y)

    flag = False

    # Traverse all possible pairs
    for i in range(1, X + 1, 1):
        for j in range(1, Y + 1, 1):
            val = (i ^ j)

            # If a pair exists
            if (val > playerA):
                flag = True
                break

        if (flag):
            break

    # If a pair is found
    if (flag):
        print("Yes")
    else:
        print("No")

# Driver Code
if __name__ == '__main__':
    
    A = 2
    B = 4
    
    findWinner(A, B)

# This code is contributed by bgangwar59

// C# program for the above approach
using System.Collections.Generic; 
using System; 

class GFG{
    
// Function to check if there exists
// any pair (P, Q) whose Bitwise XOR
// is greater than the Bitwise XOR
// of X and Y
static void findWinner(int X, int Y)
{
    
    // Stores the Bitwise XOR of X & Y
    int playerA = (X ^ Y);

    bool flag = false;

    // Traverse all possible pairs
    for(int i = 1; i <= X; i++) 
    {
        for(int j = 1; j <= Y; j++)
        {
            int val = (i ^ j);

            // If a pair exists
            if (val > playerA) 
            {
                flag = true;
                break;
            }
        }

        if (flag)
        {
            break;
        }
    }

    // If a pair is found
    if (flag) 
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}

// Driver code
public static void Main(String[] args)
{ 
    int A = 2, B = 4;
    
    findWinner(A, B);
}
}

// This code is contributed by amreshkumar3

<script>

// JavaScript program for the above approach

// Function to check if there exists
// any pair (P, Q) whose Bitwise XOR
// is greater than the Bitwise XOR
// of X and Y
function findWinner(X,Y)
{
    // Stores the Bitwise XOR of X & Y
    let playerA = (X ^ Y);
 
    let flag = false;
 
    // Traverse all possible pairs
    for(let i = 1; i <= X; i++)
    {
        for(let j = 1; j <= Y; j++)
        {
            let val = (i ^ j);
 
            // If a pair exists
            if (val > playerA)
            {
                flag = true;
                break;
            }
        }
 
        if (flag)
        {
            break;
        }
    }
 
    // If a pair is found
    if (flag)
    {
        document.write("Yes<br>");
    }
    else
    {
        document.write("No<br>");
    }
}

// Driver code
let A = 2, B = 4;

findWinner(A, B);


// This code is contributed by unknown2108

</script>
 

No

Time Complexity: O(X * Y)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized based on the following observations:

• For any two integers P and Q, the maximum Bitwise XOR value is (P + Q) which can only be found when there are no common bits between P and Q in their binary representation.
• There are two cases:
  • Case 1: If player A has two integers that produce the maximum Bitwise XOR value, then print "No".
  • Case 2: In this case, there must have some common bit between A and B such that there always exist two integers P and Q whose Bitwise XOR is always greater than the Bitwise XOR of A and B, where (P ^ Q) = (X | Y).

Therefore, from the above observations, the idea is to check if the value of given A^B is equal to A + B or not. If found to be true, then print "No". Otherwise, print "Yes".

Below is the implementation of the above approach:

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if there exists
// any pair (P, Q) whose Bitwise XOR
// is greater than the Bitwise XOR
// of X and Y
void findWinner(int X, int Y)
{
    int first = (X ^ Y);
    int second = (X + Y);

    // Check for the invalid condition
    if (first == second) {
        cout << "No";
    }

    // Otherwise,
    else {
        cout << "Yes";
    }
}

// Driver Code
int main()
{
    int A = 2, B = 4;
    findWinner(A, B);

    return 0;
}

// Java program for the above approach
import java.lang.*;
import java.util.*;

class GFG{
    
// Function to check if there exists
// any pair (P, Q) whose Bitwise XOR
// is greater than the Bitwise XOR
// of X and Y
static void findWinner(int X, int Y)
{
    int first = (X ^ Y);
    int second = (X + Y);

    // Check for the invalid condition
    if (first == second)
    {
        System.out.println("No");
    }

    // Otherwise,
    else 
    {
        System.out.println("Yes");
    }
}

// Driver code
public static void main(String[] args)
{ 
    int A = 2, B = 4;
    
    findWinner(A, B);
}
}

// This code is contributed by offbeat

# Python3 program for the above approach

# Function to check if there exists
# any pair (P, Q) whose Bitwise XOR
# is greater than the Bitwise XOR
# of X and Y
def findWinner(X, Y):
    
    first = (X ^ Y)
    second = (X + Y)

    # Check for the invalid condition
    if (first == second):
        print ("No")

    # Otherwise,
    else:
        print ("Yes")

# Driver Code
if __name__ == '__main__':
    
    A, B = 2, 4
    
    findWinner(A, B)

# This code is contributed by mohit kumar 29

// C# program for the above approach
using System;

class GFG{
    
// Function to check if there exists
// any pair (P, Q) whose Bitwise XOR
// is greater than the Bitwise XOR
// of X and Y
static void findWinner(int X, int Y)
{
    int first = (X ^ Y);
    int second = (X + Y);

    // Check for the invalid condition
    if (first == second)
    {
        Console.Write("No");
    }

    // Otherwise,
    else 
    {
        Console.Write("Yes");
    }
}

// Driver code
public static void Main(String[] args)
{ 
    int A = 2, B = 4;
    
    findWinner(A, B);
}
}

// This code is contributed by shivanisinghss2110

<script>

// JavaScript program for the above approach

// Function to check if there exists
// any pair (P, Q) whose Bitwise XOR
// is greater than the Bitwise XOR
// of X and Y
function findWinner(X,Y)
{
    let first = (X ^ Y);
    let second = (X + Y);
 
    // Check for the invalid condition
    if (first == second) {
        document.write("No");
    }
 
    // Otherwise,
    else {
        document.write("Yes");
    }
}

// Driver Code
let A = 2, B = 4;
findWinner(A, B);


// This code is contributed by patel2127

</script>

No

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