Queries to find the XOR of an Array after replacing all occurrences of X by Y

Given an array arr[] consisting of N distinct integers and queries Q[][] of the type {X, Y}, the task for each query is to find the bitwise XOR of all the array elements after replacing X by Y in the array.

Examples:

Input: arr[] = {1, 2, 3, 4, 5} Q = {(1, 4) (3, 6) (2, 3)} 
Output:4 1 0 
Explanation:
Query 1: The array modifies to {4, 2, 3, 4, 5} and XOR = 4 
Query 2: The array modifies to {4, 2, 6, 4, 5} and XOR = 1 
Query 3: The array modifies to {4, 3, 6, 4, 5} and XOR = 0
Input: arr[] = {5, 7, 8, 9, } Q = {(5, 6) (8, 1)} 
Output: 0 9 
Explanation:
Query 1: The array modifies to {6, 7, 8, 9} and XOR = 0 
Query 2: The array modifies to {6, 7, 1, 9} and XOR = 9

Approach:
The approach is to use the Bitwise XOR property:

• Suppose, there are three elements A, B, and and X, and their Xor is, total_xor = A ^ B ^ X.
• To remove the contribution of X from total_xor, perform total_xor ^ X. It can be verified from the fact that A ^ B ^ X ^ X = A ^ B (XOR of an element with itself is 0).
• To add the contribution of Y in the total_xor, simply perform total_xor ^ Y.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Stores the bitwise XOR
// of array elements
int total_xor;

// Function to find the total xor
void initialize_xor(int arr[], int n)
{
    // Loop to find the xor
    // of all the elements
    for (int i = 0; i < n; i++) {
        total_xor = total_xor ^ arr[i];
    }
}

// Function to find the XOR
// after replacing all occurrences
// of X by Y for Q queries
void find_xor(int X, int Y)
{
    // Remove contribution of
    // X from total_xor
    total_xor = total_xor ^ X;

    // Adding contribution of
    // Y to total_xor
    total_xor = total_xor ^ Y;

    // Print total_xor
    cout << total_xor << "\n";
}

// Driver Code
int main()
{
    int arr[] = { 5, 7, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);

    initialize_xor(arr, n);

    vector<vector<int> > Q = { { 5, 6 }, { 8, 1 } };

    for (int i = 0; i < Q.size(); i++) {
        find_xor(Q[i][0], Q[i][1]);
    }
    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{

// Stores the bitwise XOR
// of array elements
static int total_xor;

// Function to find the total xor
static void initialize_xor(int arr[],
                           int n)
{
    // Loop to find the xor
    // of all the elements
    for (int i = 0; i < n; i++)
    {
        total_xor = total_xor ^ arr[i];
    }
}

// Function to find the XOR
// after replacing all occurrences
// of X by Y for Q queries
static void find_xor(int X, int Y)
{
    // Remove contribution of
    // X from total_xor
    total_xor = total_xor ^ X;

    // Adding contribution of
    // Y to total_xor
    total_xor = total_xor ^ Y;

    // Print total_xor
    System.out.print(total_xor + "\n");
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 5, 7, 8, 9 };
    int n = arr.length;

    initialize_xor(arr, n);

    int [][]Q = { { 5, 6 }, { 8, 1 } };

    for (int i = 0; i < Q.length; i++)
    {
        find_xor(Q[i][0], Q[i][1]);
    }
}
}

// This code is contributed by Rohit_ranjan

Python3

# Python3 program to implement
# the above approach

# Stores the bitwise XOR
# of array elements
global total_xor
total_xor = 0

# Function to find the total xor
def initialize_xor(arr, n):

    global total_xor

    # Loop to find the xor
    # of all the elements
    for i in range(n):
        total_xor = total_xor ^ arr[i]

# Function to find the XOR
# after replacing all occurrences
# of X by Y for Q queries
def find_xor(X, Y):

    global total_xor

    # Remove contribution of
    # X from total_xor
    total_xor = total_xor ^ X

    # Adding contribution of
    # Y to total_xor
    total_xor = total_xor ^ Y

    # Print total_xor
    print(total_xor)

# Driver Code
if __name__ == '__main__':

    arr = [ 5, 7, 8, 9 ]
    n = len(arr)

    initialize_xor(arr, n)

    Q = [ [ 5, 6 ], [ 8, 1 ] ]

    # Function call
    for i in range(len(Q)):
        find_xor(Q[i][0], Q[i][1])

# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Stores the bitwise XOR
// of array elements
static int total_xor;

// Function to find the total xor
static void initialize_xor(int []arr,
                           int n)
{

    // Loop to find the xor
    // of all the elements
    for(int i = 0; i < n; i++)
    {
        total_xor = total_xor ^ arr[i];
    }
}

// Function to find the XOR
// after replacing all occurrences
// of X by Y for Q queries
static void find_xor(int X, int Y)
{

    // Remove contribution of
    // X from total_xor
    total_xor = total_xor ^ X;

    // Adding contribution of
    // Y to total_xor
    total_xor = total_xor ^ Y;

    // Print total_xor
    Console.Write(total_xor + "\n");
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 5, 7, 8, 9 };
    int n = arr.Length;

    initialize_xor(arr, n);

    int [,]Q = { { 5, 6 }, { 8, 1 } };

    for(int i = 0; i < Q.GetLength(0); i++)
    {
        find_xor(Q[i, 0], Q[i, 1]);
    }
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript program for the above approach

// Stores the bitwise XOR
// of array elements
let total_xor;

// Function to find the total xor
function initialize_xor(arr, n)
{
    // Loop to find the xor
    // of all the elements
    for (let i = 0; i < n; i++)
    {
        total_xor = total_xor ^ arr[i];
    }
}

// Function to find the XOR
// after replacing all occurrences
// of X by Y for Q queries
function find_xor(X, Y)
{
    // Remove contribution of
    // X from total_xor
    total_xor = total_xor ^ X;

    // Adding contribution of
    // Y to total_xor
    total_xor = total_xor ^ Y;

    // Print total_xor
    document.write(total_xor + "<br/>");
}

// Driver Code

    let arr = [ 5, 7, 8, 9 ];
    let n = arr.length;

    initialize_xor(arr, n);

    let Q = [[ 5, 6 ], [ 8, 1]];

    for (let i = 0; i < Q.length; i++)
    {
        find_xor(Q[i][0], Q[i][1]);
    }

</script>

Output:

0
9

Time Complexity: O(N + sizeof(Q)) 
Auxiliary Space: O(1)