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

0 0 5 minutes read

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)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!