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Find the final Array by updating given Ranges

• Difficulty Level : Expert
• Last Updated : 05 Dec, 2022

Given an array arr[] consisting of integers and an array Q[][3] consisting of M queries of the form [L, R, U], the task for each query is to xor every array element over the range [L, R] with U, After processing each query print the final array.

Examples:

Input: arr[] = {0, 0, 0, 0, 0, 0, 0}, Q[][] = {{2, 4, 1}, {0, 4, 2}, {1, 5, 3}}
Output: Initial Array : 0  0  0  0  0  0  0
Final Array : 2  1  0  0  0  3  0
Explanation: Query 1: For query {2, 4, 1} xor every index from 2 to 4 with 1 so it becomes {0, 0, 1, 1, 1, 0, 0}
Query 2: For query {0, 4, 2} xor every index from 0 to 4 with 2 so it becomes {2, 2, 1 ^ 2, 1 ^ 2, 1 ^ 2, 0, 0} which is {2, 2, 3, 3, 3, 0, 0}
Query 3: For query {1, 5, 3} xor every index from 1 to 5 with 3 so it becomes {2, 2 ^ 3, 1 ^ 2 ^ 3, 1 ^ 2 ^ 3, 1 ^ 2 ^ 3, 3, 0} which is {2, 1, 0, 0, 0, 3, 0}

Input: arr[] = {1, 2, 4, 3, 9}, Q[][] = {{0, 3, 15}, {0, 4, 15}}
Output: Initial Array : 1  2  4  3  9
Final Array : 14  13  11  12  6

Naive Approach: The basic way to solve the problem is as follows:

Iterate from L to R and xor U to all elements from arr[L] to arr[R] for each query.

Time Complexity: O(N * M) where M is the number of queries.
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following idea:

We can utilise prefix xor (similar to prefix sum) efficiently to be able to perform following queries in constant time. Idea is if we take xor of prefix[L] with U and xor of prefix[R + 1] with U again after doing prefix XOR we can observe that L to R gets updated with xor with U. But when we try to take xor at R + 1 XOR that was carrying itself forward from L in prefix xor gets deleted at R + 1 (since x ^ x == 0).

For Example: {0, 0, 0, 0, 0} and L = 1, R = 3 and U = 10

Updating array at L and R + 1 with U: {0, 10, 0, 0, 10}

• STEP 1: {0, 10, 0, 0, 10}
• STEP 2: {0, 10, 10, 0, 10}
• STEP 3: {0, 10, 10, 10, 10}
• STEP 4: {0, 10, 10, 10, 10 ^ 10} that is {0, 10, 10, 10, 0}

This can be observed that 10 was carrying itself forward but at R + 1 it gets deleted and range from L to R  gets updated with xor U. This can be used for all queries.

Follow the steps below to solve the problem:

• The pre-computation array prefix[] is defined
• Traversing for each M query and taking xor of prefix[L] with U and prefix[R + 1] with U.
• Taking prefix xor for every U for M queries carries itself forward from L to R and ends at R + 1.
• Taking xor of respective indexes of the initial array with prefix array and then printing it.

Below is the implementation of the above approach:

C++

```// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to process given range xor
// update queries then printing the
// final array
void printFinalArray(int arr[], int N, int Q[][3], int M)
{
// printing initial array
cout << "Initial Array: ";
for (int i = 0; i < N; i++) {
cout << arr[i] << "  ";
}
cout << endl;

// Precomputation array
// used for range update
int prefix[N + 1] = { 0 };

// Traverse the given queries
for (int i = 0; i < M; i++) {

// Stores range L, R and U for
// each query
int L = Q[i][0], R = Q[i][1], U = Q[i][2];

for (int i = L; i <= R; i++) {
// Updating L and R + 1 with U
prefix[L] = prefix[L] ^ U;
prefix[R + 1] = prefix[R + 1] ^ U;
}
}

// Taking prefix xor
// so that every U for M queries
// carries itself forward from L
// to R and end at R + 1
for (int i = 1; i < N; i++) {
prefix[i] = prefix[i] ^ prefix[i - 1];
}

// finally printing Final array
cout << "Final Array: ";
for (int i = 0; i < N; i++) {

// Taking xor with initial array of
// prefix array so range updates
// can be made in initial array
arr[i] = arr[i] ^ prefix[i];
cout << arr[i] << "  ";
}
}

// Driver Code
int main()
{
int arr[] = { 0, 0, 0, 0, 0, 0, 0 };
int Q[][3] = { { 2, 4, 1 }, { 0, 4, 2 }, { 1, 5, 3 } };
int N = sizeof(arr) / sizeof(arr[0]);
int M = sizeof(Q) / sizeof(Q[0]);

// Function Call
printFinalArray(arr, N, Q, M);

return 0;
}```

Java

```// Java code to implement the approach

import java.io.*;
import java.util.*;

class GFG {

// Function to process given range xor
// update queries then printing the
// final array
static void printFinalArray(int[] arr, int N, int[][] Q,
int M)
{
// Printing initial array
System.out.print("Initial Array: ");
for (int i = 0; i < N; i++) {
System.out.print(arr[i] + "  ");
}
System.out.println();

// Precomputation array
// used for range update
int[] prefix = new int[N + 1];
Arrays.fill(prefix, 0);

// Traverse the given queries
for (int i = 0; i < M; i++) {

// Stores range L, R and U for
// each query
int L = Q[i][0], R = Q[i][1], U = Q[i][2];

for (int j = L; j <= R; j++) {
// Updating L and R + 1 with U
prefix[L] = prefix[L] ^ U;
prefix[R + 1] = prefix[R + 1] ^ U;
}
}

// Taking prefix xor
// so that every U for M queries
// carries itself forward from L
// to R and end at R + 1
for (int i = 1; i < N; i++) {
prefix[i] = prefix[i] ^ prefix[i - 1];
}

// finally printing Final array
System.out.print("Final Array: ");
for (int i = 0; i < N; i++) {

// Taking xor with initial array of
// prefix array so range updates
// can be made in initial array
arr[i] = arr[i] ^ prefix[i];
System.out.print(arr[i] + "  ");
}
}

public static void main(String[] args)
{
int[] arr = { 0, 0, 0, 0, 0, 0, 0 };
int[][] Q
= { { 2, 4, 1 }, { 0, 4, 2 }, { 1, 5, 3 } };
int N = arr.length;
int M = Q.length;

// Function call
printFinalArray(arr, N, Q, M);
}
}

// This code is contributed by lokesh.```

Python3

```# Python code to implement the approach

# Function to process given range xor
# update queries then printing the
# final array
def printFinalArray(arr,N,Q,M):
# printing initial array
print("Initial Array: ")
for i in range(N):
print(arr[i],end=" ")
print()

# Precomputation array
# used for range update
prefix=[0]*(N+1)

# Traverse the given queries
for i in range(M):
# Stores range L, R and U for
# each query
L=Q[i][0]
R=Q[i][1]
U=Q[i][2]

for i in range(L,R+1):
# Updating L and R + 1 with U
prefix[L]=prefix[L]^U
prefix[R+1]=prefix[R+1]^U

# Taking prefix xor
# so that every U for M queries
# carries itself forward from L
# to R and end at R + 1
for i in range(1,N):
prefix[i]=prefix[i]^prefix[i-1]

# finally printing Final array
print("Final Array: ")
for i in range(N):
# Taking xor with initial array of
# prefix array so range updates
# can be made in initial array
arr[i]=arr[i]^prefix[i]
print(arr[i],end=" ")

# Driver Code
arr=[0, 0, 0, 0, 0, 0, 0]
Q=[[2,4,1],[0,4,2],[1,5,3]]
N=len(arr)
M=len(Q)

# Function Call
printFinalArray(arr,N,Q,M)

# This code is contributed by Pushpesh Raj.
```

C#

```// C# code to implement the approach
using System;
using System.Collections;
using System.Collections.Generic;

public class GFG {

// Function to process given range xor
// update queries then printing the
// final array
static void printFinalArray(int[] arr, int N, int[, ] Q,
int M)
{
// Printing initial array
Console.Write("Initial Array: ");
for (int i = 0; i < N; i++) {
Console.Write(arr[i] + "  ");
}
Console.WriteLine();

// Precomputation array
// used for range update
int[] prefix = new int[N + 1];
Array.Fill(prefix, 0);

// Traverse the given queries
for (int i = 0; i < M; i++) {

// Stores range L, R and U for
// each query
int L = Q[i, 0], R = Q[i, 1], U = Q[i, 2];

for (int j = L; j <= R; j++) {
// Updating L and R + 1 with U
prefix[L] = prefix[L] ^ U;
prefix[R + 1] = prefix[R + 1] ^ U;
}
}

// Taking prefix xor
// so that every U for M queries
// carries itself forward from L
// to R and end at R + 1
for (int i = 1; i < N; i++) {
prefix[i] = prefix[i] ^ prefix[i - 1];
}

// finally printing Final array
Console.Write("Final Array: ");
for (int i = 0; i < N; i++) {

// Taking xor with initial array of
// prefix array so range updates
// can be made in initial array
arr[i] = arr[i] ^ prefix[i];
Console.Write(arr[i] + "  ");
}
}

static public void Main()
{

// Code
int[] arr = { 0, 0, 0, 0, 0, 0, 0 };
int[, ] Q
= { { 2, 4, 1 }, { 0, 4, 2 }, { 1, 5, 3 } };
int N = arr.Length;
int M = Q.GetLength(0);

// Function call
printFinalArray(arr, N, Q, M);
}
}

// This code is contributed by lokeshmvs21.```

Javascript

```        // JavaScript code to implement the approach

// Function to process given range xor
// update queries then printing the
// final array
const printFinalArray = (arr, N, Q, M) => {
// printing initial array
document.write("Initial Array: ");
for (let i = 0; i < N; i++) {
console.log(`\${arr[i]} `);
}
console.log("<br/>");

// Precomputation array
// used for range update
let prefix = new Array(N + 1).fill(0);

// Traverse the given queries
for (let i = 0; i < M; i++) {

// Stores range L, R and U for
// each query
let L = Q[i][0], R = Q[i][1], U = Q[i][2];

for (let i = L; i <= R; i++) {
// Updating L and R + 1 with U
prefix[L] = prefix[L] ^ U;
prefix[R + 1] = prefix[R + 1] ^ U;
}
}

// Taking prefix xor
// so that every U for M queries
// carries itself forward from L
// to R and end at R + 1
for (let i = 1; i < N; i++) {
prefix[i] = prefix[i] ^ prefix[i - 1];
}

// finally printing Final array
console.log("Final Array: ");
for (let i = 0; i < N; i++) {

// Taking xor with initial array of
// prefix array so range updates
// can be made in initial array
arr[i] = arr[i] ^ prefix[i];
console.log(`\${arr[i]} `);
}
}

// Driver Code
let arr = [0, 0, 0, 0, 0, 0, 0];
let Q = [[2, 4, 1], [0, 4, 2], [1, 5, 3]];
let N = arr.length;
let M = Q.length;

// Function Call
printFinalArray(arr, N, Q, M);

// This code is contributed by rakeshsahni

```
Output

```Initial Array : 0  0  0  0  0  0  0
Final Array : 2  1  0  0  0  3  0  ```

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

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