Queries to update each element in subarray to Bitwise XOR with a given value

Given an array arr[], and queries Q[][] of the form (l, r, val), the task for each query is to update all the elements in the indices [l – 1, r – 1] to Bitwise XOR with val. Print the final array obtained after completing all queries.
Examples:

Input: arr[] = {2, 3, 6, 5, 4}, Q[][] = {{1, 3, 2}, {2, 4, 4}} 
Output: 0 5 0 1 4 
Explanation:
1^st Query: Concerned subarray {2, 3, 6} modifies to {0, 1, 4} after replacing each element with its XOR with val(= 2) 
The modified array is {0, 1, 4, 5, 4} 
2^nd Query: Concerned subarray {1, 4, 5} modifies to {5, 0, 1} after replacing each element with its XOR with val(= 4) 
Hence, the final array is {0, 5, 0, 1, 4}
Input: arr[] = {1, 3, 5}, Q[][] = {{1, 2, 8}, {2, 3, 3}} 
Output: 9 8 6 

Naive Approach:
The simplest approach to solve this problem is to traverse indices [l – 1, r – 1] for each query and replace arr[i] by arr[i]^val. After completing all queries, print the modified array. 
Time Complexity: O(N*sizeof(Q)) 
Auxiliary Space: O(1)
Approach: Follow the steps to solve the problem by processing each query in constant time complexity: 

• Initialize an array temp[] with all zeroes.
• For each query of the form (l, r, val), update temp[l – 1] and temp[r] with their respective XOR with val.
• After completing the above step for each query, convert temp[] to a prefix XOR array.
• Finally, perform update arr[] by replacing every i^th element with its XOR.

Below is the implementation of the above approach:

C++

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

// Function to perform XOR in
// the range [lo, hi]
void findxor(int temp[], int N,
             int lo, int hi, int val)
{
    temp[lo] ^= val;
    if (hi != N - 1)
        temp[hi + 1] ^= val;
}

// Function to generate Prefix
// Xor Array
void updateArray(int temp[], int N)
{
    for (int i = 1; i < N; i++)
        temp[i] ^= temp[i - 1];
}

// Function to perform each Query
// and print the final array
void xorQuery(int arr[], int n,
              vector<vector<int> > Q)
{

    int temp[n];
    // initialize temp array to 0
    memset(temp, 0, sizeof(temp));

    // Perform each Query
    for (int i = 0; i < Q.size(); i++) {
        findxor(temp, n, Q[i][0] - 1,
                Q[i][1] - 1, Q[i][2]);
    }

    // Modify the array
    updateArray(temp, n);

    // Print the final array
    for (int i = 0; i < n; ++i) {
        cout << (arr[i] ^ temp[i]) << " ";
    }
}

// Driver Code
int main()
{

    int arr[] = { 2, 3, 6, 5, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    vector<vector<int> > Q = { { 1, 3, 2 },
                               { 2, 4, 4 } };

    xorQuery(arr, n, Q);

    return 0;
}

Java

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

class GFG{

// Function to perform XOR in
// the range [lo, hi]
static void findxor(int temp[], int N,
                    int lo, int hi, int val)
{
    temp[lo] ^= val;
    if (hi != N - 1)
        temp[hi + 1] ^= val;
}

// Function to generate Prefix
// Xor Array
static void updateArray(int temp[], int N)
{
    for(int i = 1; i < N; i++)
        temp[i] ^= temp[i - 1];
}

// Function to perform each Query
// and print the final array
static void xorQuery(int arr[], int n,
                     int[][] Q)
{
    int[] temp = new int[n];

    // Perform each Query
    for(int i = 0; i < Q.length; i++)
    {
        findxor(temp, n, Q[i][0] - 1,
                         Q[i][1] - 1,
                         Q[i][2]);
    }

    // Modify the array
    updateArray(temp, n);

    // Print the final array
    for(int i = 0; i < n; ++i)
    {
        System.out.print((arr[i] ^ temp[i]) + " ");
    }
}

// Driver Code
public static void main (String[] args)
{
    int arr[] = { 2, 3, 6, 5, 4 };
    int n = arr.length;

    int[][] Q = { { 1, 3, 2 },
                  { 2, 4, 4 } };

    xorQuery(arr, n, Q);
}
}

// This code is contributed by offbeat

Python3

# Python3 program to implement
# the above approach

# Function to perform XOR in
# the range [lo, hi]
def findxor(temp, N, lo, hi, val):

    temp[lo] ^= val
    if (hi != N - 1):
        temp[hi + 1] ^= val

# Function to generate Prefix
# Xor Array
def updateArray(temp, N):

    for i in range(1, N):
        temp[i] ^= temp[i - 1]

# Function to perform each Query
# and print the final array
def xorQuery(arr, n, Q):

    temp =[0] * n

    # Perform each Query
    for i in range(len(Q)):
        findxor(temp, n, Q[i][0] - 1,
                         Q[i][1] - 1,
                         Q[i][2])

    # Modify the array
    updateArray(temp, n)

    # Print the final array
    for i in range(n):
        print((arr[i] ^ temp[i]), end = " ")

# Driver Code
if __name__ == "__main__":

    arr = [ 2, 3, 6, 5, 4 ]
    n = len(arr)
    Q = [ [ 1, 3, 2 ],
          [ 2, 4, 4 ] ]

    xorQuery(arr, n, Q)

# This code is contributed by chitranayal

C#

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

class GFG{

// Function to perform XOR in
// the range [lo, hi]
static void findxor(int []temp, int N,
                    int lo, int hi, int val)
{
    temp[lo] ^= val;

    if (hi != N - 1)
        temp[hi + 1] ^= val;
}

// Function to generate Prefix
// Xor Array
static void updateArray(int []temp, int N)
{
    for(int i = 1; i < N; i++)
        temp[i] ^= temp[i - 1];
}

// Function to perform each Query
// and print the readonly array
static void xorQuery(int []arr, int n,
                     int[,] Q)
{
    int[] temp = new int[n];

    // Perform each Query
    for(int i = 0; i < Q.GetLength(0); i++)
    {
        findxor(temp, n, Q[i, 0] - 1,
                         Q[i, 1] - 1,
                         Q[i, 2]);
    }

    // Modify the array
    updateArray(temp, n);

    // Print the readonly array
    for(int i = 0; i < n; ++i)
    {
        Console.Write((arr[i] ^ temp[i]) + " ");
    }
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 2, 3, 6, 5, 4 };
    int n = arr.Length;

    int[,] Q = { { 1, 3, 2 },
                 { 2, 4, 4 } };

    xorQuery(arr, n, Q);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript Program to implement
// the above approach

// Function to perform XOR in
// the range [lo, hi]
function findxor(temp, N, lo, hi, val) {
    temp[lo] ^= val;
    if (hi != N - 1)
        temp[hi + 1] ^= val;
}

// Function to generate Prefix
// Xor Array
function updateArray(temp, N) {
    for (let i = 1; i < N; i++)
        temp[i] ^= temp[i - 1];
}

// Function to perform each Query
// and print the final array
function xorQuery(arr, n, Q) {

    let temp = new Array(n);
    // initialize temp array to 0
    temp.fill(0)

    // Perform each Query
    for (let i = 0; i < Q.length; i++) {
        findxor(temp, n, Q[i][0] - 1,
            Q[i][1] - 1, Q[i][2]);
    }

    // Modify the array
    updateArray(temp, n);

    // Print the final array
    for (let i = 0; i < n; ++i) {
        document.write(`${arr[i] ^ temp[i]} `);
    }
}

// Driver Code

let arr = [2, 3, 6, 5, 4];
let n = arr.length;
let Q = [[1, 3, 2],
[2, 4, 4]];

xorQuery(arr, n, Q);

// This code is contributed by gfgking
</script>

Output:

0 5 0 1 4

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