XOR of a subarray (range of elements) | Set 2

Given an array integer arr[] of size N and Q queries. Each query is of the form (L, R), where L and R are indices of the array. The task is to find the XOR value of the subarray arr[L…R]. 

Examples:

Input: arr[] = {2, 5, 1, 6, 1, 2, 5} query[] = {{1, 4}} 
Output: 3 
The XOR value of arr[1…4] is 3.

Input: arr[] = {2, 5, 1, 6, 1, 2, 5} query[] = {{0, 6}} 
Output: 6 
The XOR value of arr[0…6] is 6.  

O(N) auxiliary space approach: Please refer to this article for the O(N) auxiliary space approach.

Approach: Here we will discuss a constant space solution, but we will modify the input array.  

1. Update the input array from index 1 to N – 1, such that arr[i] stores the XOR from arr[0] to arr[i]. 

arr[i] = XOR(arr[0], arr[1], .., arr[i]) 

2. To process a query from L to R just return arr[L-1] ^ arr[R].

For example:

Consider the example: arr[] = { 3, 2, 4, 5, 1, 1, 5, 3 }, query[] = {{1, 4 }, { 3, 7}} 
After taking the XOR of consecutive elements, arr[] = {3, 1, 5, 0, 1, 0, 5, 6} 
For first query {1, 4} ans = arr[0] ^ arr[4] = 3 ^ 1 = 2 
For the second query {3, 7} ans = arr[2] ^ arr[7] = 5 ^ 6 = 3  

Below is the implementation of the above approach: 

C++

// C++ program to find XOR
// in a range from L to R

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

// Function to find XOR
// in a range from L to R
void find_Xor(int arr[],
              pair<int, int> query[],
              int N, int Q)
{
    // Compute xor from arr[0] to arr[i]
    for (int i = 1; i < N; i++) {
        arr[i] = arr[i] ^ arr[i - 1];
    }

    int ans = 0;

    // process every query
    // in constant time
    for (int i = 0; i < Q; i++) {

        // if L==0
        if (query[i].first == 0)
            ans = arr[query[i].second];
        else
            ans = arr[query[i].first - 1]
                  ^ arr[query[i].second];

        cout << ans << endl;
    }
}

// Driver Code
int main()
{
    int arr[] = { 3, 2, 4, 5,
                  1, 1, 5, 3 };
    int N = 8;
    int Q = 2;
    pair<int, int> query[Q]
        = { { 1, 4 },
            { 3, 7 } };

    // query[]
    find_Xor(arr, query, N, Q);
    return 0;
}

Java

// Java program to find XOR
// in a range from L to R
class GFG{

static class pair
{
    int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function to find XOR
// in a range from L to R
static void find_Xor(int arr[],
                     pair query[],
                     int N, int Q)
{

    // Compute xor from arr[0] to arr[i]
    for(int i = 1; i < N; i++)
    {
       arr[i] = arr[i] ^ arr[i - 1];
    }

    int ans = 0;

    // Process every query
    // in constant time
    for(int i = 0; i < Q; i++)
    {

       // If L==0
       if (query[i].first == 0)
           ans = arr[query[i].second];
       else
           ans = arr[query[i].first - 1] ^
                 arr[query[i].second];

       System.out.print(ans + "\n");
    }
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 4, 5,
                  1, 1, 5, 3 };
    int N = 8;
    int Q = 2;

    pair query[] = { new pair(1, 4),
                     new pair(3, 7) };

    // query[]
    find_Xor(arr, query, N, Q);
}
}

// This code is contributed by gauravrajput1

Python3

# Python3 program to find XOR
# in a range from L to R

# Function to find XOR
# in a range from L to R
def find_Xor(arr, query, N, Q):

    # Compute xor from arr[0] to arr[i]
    for i in range(1, N):
        arr[i] = arr[i] ^ arr[i - 1]

    ans = 0

    # Process every query
    # in constant time
    for i in range(Q):

        # If L == 0
        if query[i][0] == 0:
            ans = arr[query[i][1]]
        else:
            ans = (arr[query[i][0] - 1] ^
                   arr[query[i][1]])
        print(ans)

# Driver code
def main():

    arr = [ 3, 2, 4, 5, 1, 1, 5, 3 ]
    N = 8
    Q = 2

    # query[]
    query = [ [ 1, 4 ],
              [ 3, 7 ] ]

    find_Xor(arr, query, N, Q)

main()

# This code is contributed by Stuti Pathak

C#

// C# program to find XOR
// in a range from L to R
using System;

class GFG{
class pair
{
    public int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function to find XOR
// in a range from L to R
static void find_Xor(int []arr,
                     pair []query,
                     int N, int Q)
{

    // Compute xor from arr[0] to arr[i]
    for(int i = 1; i < N; i++)
    {
       arr[i] = arr[i] ^ arr[i - 1];
    }

    int ans = 0;

    // Process every query
    // in constant time
    for(int i = 0; i < Q; i++)
    {

       // If L==0
       if (query[i].first == 0)
           ans = arr[query[i].second];
       else
           ans = arr[query[i].first - 1] ^
                 arr[query[i].second];

       Console.Write(ans + "\n");
    }
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 3, 2, 4, 5,
                  1, 1, 5, 3 };
    int N = 8;
    int Q = 2;

    pair []query = { new pair(1, 4),
                     new pair(3, 7) };

    // query[]
    find_Xor(arr, query, N, Q);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript program to find XOR
// in a range from L to R

// Function to find XOR
// in a range from L to R
function find_Xor(arr, query, N, Q)
{
    // Compute xor from arr[0] to arr[i]
    for (var i = 1; i < N; i++) {
        arr[i] = arr[i] ^ arr[i - 1];
    }

    var ans = 0;

    // process every query
    // in constant time
    for (var i = 0; i < Q; i++) {

        // if L==0
        if (query[i][0] == 0)
            ans = arr[query[i][1]];
        else
            ans = arr[query[i][0] - 1]
                  ^ arr[query[i][1]];

        document.write( ans + "<br>");
    }
}

// Driver Code
var arr = [3, 2, 4, 5,
              1, 1, 5, 3];
var N = 8;
var Q = 2;
var query
    = [[1, 4],
        [3, 7]];
// query[]
find_Xor(arr, query, N, Q);


</script>

Output:

2
3

Time Complexity: O (N + Q)
Auxiliary Space: O (1)