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)}Â
Output:4 1 0Â
Explanation:Â
Query 1: The array modifies to {4, 2, 3, 4, 5} and XOR = 4Â
Query 2: The array modifies to {4, 2, 6, 4, 5} and XOR = 1Â
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)}Â
Output: 0 9Â
Explanation:Â
Query 1: The array modifies to {6, 7, 8, 9} and XOR = 0Â
Query 2: The array modifies to {6, 7, 1, 9} and XOR = 9
Approach:Â
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;Â
// Stores the bitwise XOR// of array elementsint total_xor;Â
// Function to find the total xorvoid 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];    }}Â
// Function to find the XOR// after replacing all occurrences// of X by Y for Q queriesvoid find_xor(int X, int Y){    // Remove contribution of    // X from total_xor    total_xor = total_xor ^ X;Â
    // Adding contribution of    // Y to total_xor    total_xor = total_xor ^ Y;Â
    // Print total_xor    cout << total_xor << "\n";}Â
// Driver Codeint main(){Â Â Â Â int arr[] = { 5, 7, 8, 9 };Â Â Â Â int n = sizeof(arr) / sizeof(arr[0]);Â
    initialize_xor(arr, n);Â
    vector<vector<int> > Q = { { 5, 6 }, { 8, 1 } };Â
    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 approachimport java.util.*;class GFG{Â
// Stores the bitwise XOR// of array elementsstatic int total_xor;Â
// Function to find the total xorstatic 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];    }}Â
// Function to find the XOR// after replacing all occurrences// of X by Y for Q queriesstatic void find_xor(int X, int Y){    // Remove contribution of    // X from total_xor    total_xor = total_xor ^ X;Â
    // Adding contribution of    // Y to total_xor    total_xor = total_xor ^ Y;Â
    // Print total_xor    System.out.print(total_xor + "\n");}Â
// Driver Codepublic static void main(String[] args){Â Â Â Â int arr[] = { 5, 7, 8, 9 };Â Â Â Â int n = arr.length;Â
    initialize_xor(arr, n);Â
    int [][]Q = { { 5, 6 }, { 8, 1 } };Â
    for (int i = 0; i < Q.length; i++)    {        find_xor(Q[i][0], Q[i][1]);    }}}Â
// This code is contributed by Rohit_ranjan |
Python3
# Python3 program to implement# the above approachÂ
# Stores the bitwise XOR# of array elementsglobal total_xortotal_xor = 0Â
# Function to find the total xordef initialize_xor(arr, n):Â
    global total_xorÂ
    # Loop to find the xor    # of all the elements    for i in range(n):        total_xor = total_xor ^ arr[i]Â
# Function to find the XOR# after replacing all occurrences# of X by Y for Q queriesdef find_xor(X, Y):Â Â Â Â Â Â Â Â Â global total_xorÂ
    # Remove contribution of    # X from total_xor    total_xor = total_xor ^ XÂ
    # Adding contribution of    # Y to total_xor    total_xor = total_xor ^ YÂ
    # Print total_xor     print(total_xor)Â
# Driver Codeif __name__ == '__main__':Â
    arr = [ 5, 7, 8, 9 ]    n = len(arr)Â
    initialize_xor(arr, n)Â
    Q = [ [ 5, 6 ], [ 8, 1 ] ]Â
    # Function call    for i in range(len(Q)):        find_xor(Q[i][0], Q[i][1])Â
# This code is contributed by Shivam Singh |
C#
// C# program to implement// the above approachusing System;Â
class GFG{Â
// Stores the bitwise XOR// of array elementsstatic int total_xor;Â
// Function to find the total xorstatic 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];    }}Â
// Function to find the XOR// after replacing all occurrences// of X by Y for Q queriesstatic void find_xor(int X, int Y){         // Remove contribution of    // X from total_xor    total_xor = total_xor ^ X;Â
    // Adding contribution of    // Y to total_xor    total_xor = total_xor ^ Y;Â
    // Print total_xor    Console.Write(total_xor + "\n");}Â
// Driver Codepublic static void Main(String[] args){Â Â Â Â int []arr = { 5, 7, 8, 9 };Â Â Â Â int n = arr.Length;Â
    initialize_xor(arr, n);Â
    int [,]Q = { { 5, 6 }, { 8, 1 } };Â
    for(int i = 0; i < Q.GetLength(0); i++)    {        find_xor(Q[i, 0], Q[i, 1]);    }}}Â
// This code is contributed by 29AjayKumar |
Javascript
<script>Â
// JavaScript program for the above approachÂ
// Stores the bitwise XOR// of array elementslet total_xor;   // Function to find the total xorfunction 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 queriesfunction 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:Â
0 9
Â
Time Complexity: O(N + sizeof(Q))Â
Auxiliary Space: O(1)
Â
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 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
