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 elementsint total_xor;// 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]; }}// 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; // Adding contribution of // Y to total_xor total_xor = total_xor ^ Y; // Print total_xor cout << total_xor << "\n";}// Driver Codeint 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 approachimport java.util.*;class GFG{// Stores the bitwise XOR// of array elementsstatic int total_xor;// 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]; }}// 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; // Adding contribution of // Y to total_xor total_xor = total_xor ^ Y; // Print total_xor System.out.print(total_xor + "\n");}// Driver Codepublic 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 elementsglobal total_xortotal_xor = 0# Function to find the total xordef 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 queriesdef 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 Codeif __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 approachusing System;class GFG{// Stores the bitwise XOR// of array elementsstatic int total_xor;// 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]; }}// 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; // Adding contribution of // Y to total_xor total_xor = total_xor ^ Y; // Print total_xor Console.Write(total_xor + "\n");}// Driver Codepublic 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 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:
0 9
Time Complexity: O(N + sizeof(Q))
Auxiliary Space: O(1)
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