Find the XOR of the elements in the given range [L, R] with the value K for a given set of queries

Given an array arr[] and Q queries, the task is to find the resulting updated array after Q queries. There are two types of queries and the following operation is performed by them: 

1. Update(L, R, K): Perform XOR for each element within the range L to R with K.
2. Display(): Display the current state of the given array.

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6}, Q = 2 
Query 1: Update(1, 5, 3) 
Query 2: Display() 
Output: 2 1 0 7 6 6 
Explanation: 
The update query performs XOR of all elements in the range [1, 5]. 
After performing the update query, the above array is displayed for display query because: 
1 ^ 3 = 2 
2 ^ 3 = 1 
3 ^ 3 = 0 
4 ^ 3 = 7 
5 ^ 3 = 6
Input: arr[] = {2, 4, 6, 8, 10}, Q = 3 
Query 1: Update(1, 3, 2) 
Query 2: Update(2, 4, 3) 
Query 3: Display() 
Output: 0 5 7 11 10 
Explanation:
There are two update queries. 
After performing the first update query, the given array is changed to {0, 6, 4, 8, 10}. This is because: 
2 ^ 2 = 0 
4 ^ 2 = 6 
6 ^ 2 = 4 
After obtaining this array, this array further gets changed for the second query as the {0, 5, 7, 11, 10}. This is because: 
6 ^ 3 = 5 
4 ^ 3 = 7 
8 ^ 3 = 11 

Naive Approach: The naive approach for this problem is for every update query, run a loop from L to R and perform the XOR operation with the given K with all the elements from arr[L] to arr[R]. In order to perform the display query, simply print the array arr[]. The time complexity for this approach is O(NQ) where N is the length of the array and Q is the number of queries. 
Efficient Approach: The idea is to use a kadane’s algorithm for this problem. 

• An empty array res[] is defined with the same length as the original array.
• For every update query {L, R, K}, the res[] array is modified as: 
  1. XOR K to res[L]
  2. XOR K to res[R + 1]
• Whenever the user gives a display query: 
  1. Initialize a counter variable ‘i’ to the index 1.
  2. Perform the operation: 
     

res[i] = res[i] ^ res[i - 1]

1. Initialize a counter variable ‘i’ to the index 0.
2. For every index in the array’s the required answer is:

arr[i] ^ res[i]

Below is the implementation of the above approach:

C++

// C++ implementation to perform the
// XOR range updates on an array
#include <bits/stdc++.h>
using namespace std;

// Function to perform the update operation
// on the given array
void update(int res[], int L, int R, int K)
{

    // Converting the indices to
    // 0 indexing.
    L -= 1;
    R -= 1;

    // Saving the XOR of K from the starting
    // index in the range [L, R].
    res[L] ^= K;

    // Saving the XOR of K at the ending index
    // in the given [L, R].
    res[R + 1] ^= K;
}

// Function to display the resulting array
void display(int arr[], int res[], int n)
{
    for (int i = 1; i < n; i++) {

        // Finding the resultant value in the
        // result array
        res[i] = res[i] ^ res[i - 1];
    }

    for (int i = 0; i < n; i++) {

        // Combining the effects of the updates
        // with the original array without
        // changing the initial array.
        cout << (arr[i] ^ res[i]) << " ";
    }
    cout << endl;
}

// Driver code
int main()
{
    int arr[] = { 2, 4, 6, 8, 10 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int res[N];
    memset(res, 0, sizeof(res));

    // Query 1
    int L = 1, R = 3, K = 2;
    update(res, L, R, K);

    // Query 2
    L = 2;
    R = 4;
    K = 3;
    update(res, L, R, K);

    // Query 3
    display(arr, res, N);

    return 0;
}

Java

// Java implementation to perform the
// XOR range updates on an array
class GFG{

// Function to perform the update
// operation on the given array
static void update(int res[], int L,
                   int R, int K)
{

    // Converting the indices to
    // 0 indexing.
    L -= 1;
    R -= 1;

    // Saving the XOR of K from the
    // starting index in the range [L, R].
    res[L] ^= K;

    // Saving the XOR of K at the ending
    // index in the given [L, R].
    res[R + 1] ^= K;
}

// Function to display the resulting array
static void display(int arr[],
                    int res[], int n)
{
    int i;
    for(i = 1; i < n; i++)
    {

       // Finding the resultant value
       // in the result array
       res[i] = res[i] ^ res[i - 1];
    }

    for(i = 0; i < n; i++)
    {

       // Combining the effects of the
       // updates with the original array
       // without changing the initial array.
       System.out.print((arr[i] ^ res[i]) + " ");
    }
    System.out.println();
}

// Driver code
public static void main(String []args)
{
    int arr[] = { 2, 4, 6, 8, 10 };
    int N = arr.length;
    int res[] = new int[N];

    // Query 1
    int L = 1, R = 3, K = 2;
    update(res, L, R, K);

    // Query 2
    L = 2;
    R = 4;
    K = 3;
    update(res, L, R, K);

    // Query 3
    display(arr, res, N);
}
}

// This code is contributed by chitranayal

Python3

# Python3 implementation to perform the
# XOR range updates on an array

# Function to perform the update operation
# on the given array
def update(res, L, R, K):

    # Converting the indices to
    # 0 indexing.
    L = L - 1
    R = R - 1

    # Saving the XOR of K from the starting
    # index in the range [L, R].
    res[L] = res[L] ^ K

    # Saving the XOR of K at the ending index
    # in the given [L, R].
    res[R + 1] = res[R + 1] ^ K

# Function to display the resulting array
def display(arr, res, n):

    for i in range(1, n):

        # Finding the resultant value in the
        # result array
        res[i] = res[i] ^ res[i - 1]

    for i in range(0, n):

        # Combining the effects of the updates
        # with the original array without
        # changing the initial array.
        print (arr[i] ^ res[i], end = " ")

# Driver code
arr = [ 2, 4, 6, 8, 10 ]
N = len(arr)
res = [0] * N

# Query 1
L = 1
R = 3
K = 2
update(res, L, R, K)

# Query 2
L = 2
R = 4
K = 3
update(res, L, R, K)

# Query 3
display(arr, res, N)

# This code is contributed by Pratik Basu

C#

// C# implementation to perform the
// XOR range updates on an array
using System;
class GFG{

// Function to perform the update
// operation on the given array
static void update(int []res, int L,
                   int R, int K)
{

    // Converting the indices to
    // 0 indexing.
    L -= 1;
    R -= 1;

    // Saving the XOR of K from the
    // starting index in the range [L, R].
    res[L] ^= K;

    // Saving the XOR of K at the ending
    // index in the given [L, R].
    res[R + 1] ^= K;
}

// Function to display the resulting array
static void display(int []arr,
                    int []res, int n)
{
    int i;
    for(i = 1; i < n; i++)
    {

        // Finding the resultant value
        // in the result array
        res[i] = res[i] ^ res[i - 1];
    }

    for(i = 0; i < n; i++)
    {

        // Combining the effects of the
        // updates with the original array
        // without changing the initial array.
        Console.Write((arr[i] ^ res[i]) + " ");
    }
    Console.WriteLine();
}

// Driver code
public static void Main()
{
    int []arr = { 2, 4, 6, 8, 10 };
    int N = arr.Length;
    int []res = new int[N];

    // Query 1
    int L = 1, R = 3, K = 2;
    update(res, L, R, K);

    // Query 2
    L = 2;
    R = 4;
    K = 3;
    update(res, L, R, K);

    // Query 3
    display(arr, res, N);
}
}

// This code is contributed by Code_Mech

Javascript

<script>

// Javascript implementation to perform the
// XOR range updates on an array

// Function to perform the update
// operation on the given array
function update(res, L, R, K)
{

    // Converting the indices to
    // 0 indexing.
    L -= 1;
    R -= 1;

    // Saving the XOR of K from the
    // starting index in the range [L, R].
    res[L] ^= K;

    // Saving the XOR of K at the ending
    // index in the given [L, R].
    res[R + 1] ^= K;
}

// Function to display the resulting array
function display(arr, res, n)
{
    let i;
    for(i = 1; i < n; i++)
    {

       // Finding the resultant value
       // in the result array
       res[i] = res[i] ^ res[i - 1];
    }

    for(i = 0; i < n; i++)
    {

       // Combining the effects of the
       // updates with the original array
       // without changing the initial array.
       document.write((arr[i] ^ res[i]) + " ");
    }
    document.write( "<br/>");
}


// Driver Code

    let arr = [ 2, 4, 6, 8, 10 ];
    let N = arr.length;
    let res = Array.from({length: N}, (_, i) => 0);

    // Query 1
    let L = 1, R = 3, K = 2;
    update(res, L, R, K);

    // Query 2
    L = 2;
    R = 4;
    K = 3;
    update(res, L, R, K);

    // Query 3
    display(arr, res, N);

</script>

Output:

0 5 7 11 10

Time Complexity:

• The time complexity for the update is O(1).
• The time complexity for displaying the array is O(N).

Auxiliary Space: O(N)

Note: This approach works very well when the update queries are very high compared to the display queries.