Length of the longest subarray whose Bitwise XOR is K

Given an array arr[] of size N and an integer K, the task is to find the length of the longest subarray having Bitwise XOR of all its elements equal to K.

Examples:

Input: arr[] = { 1, 2, 4, 7, 2 }, K = 1
Output: 3
Explanation:
Subarray having Bitwise XOR equal to K(= 1) are { { 1 }, { 2, 4, 7 }, { 1 } }.
Therefore, the length of longest subarray having bitwise XOR equal to K(= 1) is 3

Input: arr[] = { 2, 5, 6, 1, 0, 3, 5, 6 }, K = 4
Output: 6
Explanation:
Subarray having Bitwise XOR equal to K(= 4) are { { 6, 1, 0, 3 }, { 5, 6, 1, 0, 3, 5 } }.
Therefore, the length of longest subarray having bitwise XOR equal to K(= 4) is 6.

Approach: The problem can be solved using Hashing and Prefix Sum technique. Following are the observation:

a₁ ^ a₂ ^ a₃ ^ ….. ^ a_n = K

=> a₂ ^ a₃ ^ ….. ^ a_n ^ K = a₁

Follow the steps below to solve the problem:

• Initialize a variable, say prefixXOR, to store the Bitwise XOR of all elements up to the i^th index of the given array.
• Initialize a Map, say mp, to store the indices of the computed prefix XORs of the array.
• Initialize a variable, say maxLen, to store the length of the longest subarray whose Bitwise XOR is equal to K.
• Traverse the array arr[] using variable i. For every i^th index, update prefixXOR = prefixXOR ^ arr[i] and check if (prefixXOR ^ K) is present in the Map or not. If found to be true, then update maxLen = max(maxLen, i – mp[prefixXOR ^ K]).
• If prefixXOR is not present in the Map, then insert prefixXOR into the Map.
• Finally, print the value of maxLen.

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 find the length of the longest
// subarray whose bitwise XOR is equal to K
int LongestLenXORK(int arr[], int N, int K)
{

    // Stores prefix XOR
    // of the array
    int prefixXOR = 0;

    // Stores length of longest subarray
    // having bitwise XOR equal to K
    int maxLen = 0;

    // Stores index of prefix
    // XOR of the array
    unordered_map<int, int> mp;


    // Insert 0 into the map
    mp[0] = -1;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // Update prefixXOR
        prefixXOR ^= arr[i];

        // If (prefixXOR ^ K) present
        // in the map
        if (mp.count(prefixXOR ^ K)) {

            // Update maxLen
            maxLen = max(maxLen,
               (i - mp[prefixXOR ^ K]));
        }

        // If prefixXOR not present
        // in the Map
        if (!mp.count(prefixXOR)) {

            // Insert prefixXOR
            // into the map
            mp[prefixXOR] = i;
        }
    }

    return maxLen;
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 4, 7, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 1;
    cout<< LongestLenXORK(arr, N, K);

    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG{

// Function to find the length of the longest
// subarray whose bitwise XOR is equal to K
static int LongestLenXORK(int arr[],
                          int N, int K)
{

    // Stores prefix XOR
    // of the array
    int prefixXOR = 0;

    // Stores length of longest subarray
    // having bitwise XOR equal to K
    int maxLen = 0;

    // Stores index of prefix
    // XOR of the array
    HashMap<Integer,
            Integer> mp = new HashMap<Integer,
                                      Integer>();

    // Insert 0 into the map
    mp.put(0, -1);

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // Update prefixXOR
        prefixXOR ^= arr[i];

        // If (prefixXOR ^ K) present
        // in the map
        if (mp.containsKey(prefixXOR ^ K))
        {

            // Update maxLen
            maxLen = Math.max(maxLen,
               (i - mp.get(prefixXOR ^ K)));
        }

        // If prefixXOR not present
        // in the Map
        if (!mp.containsKey(prefixXOR))
        {

            // Insert prefixXOR
            // into the map
            mp.put(prefixXOR, i);
        }
    }
    return maxLen;
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 4, 7, 2 };
    int N = arr.length;
    int K = 1;

    System.out.print(LongestLenXORK(arr, N, K));
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement
# the above approach

# Function to find the length of the longest
# subarray whose bitwise XOR is equal to K
def LongestLenXORK(arr, N, K):

    # Stores prefix XOR
    # of the array
    prefixXOR = 0

    # Stores length of longest subarray
    # having bitwise XOR equal to K
    maxLen = 0

    # Stores index of prefix
    # XOR of the array
    mp = {}

    # Insert 0 into the map
    mp[0] = -1

    # Traverse the array
    for i in range(N):

        # Update prefixXOR
        prefixXOR ^= arr[i]

        # If (prefixXOR ^ K) present
        # in the map
        if (prefixXOR ^ K) in mp:

            # Update maxLen
            maxLen = max(maxLen,
                        (i - mp[prefixXOR ^ K]))

        # If prefixXOR not present
        # in the Map
        else:

            # Insert prefixXOR
            # into the map
            mp[prefixXOR] = i

    return maxLen

# Driver Code
if __name__ == "__main__" :

    arr = [ 1, 2, 4, 7, 2 ]
    N = len(arr)
    K = 1

    print(LongestLenXORK(arr, N, K))

# This code is contributed by AnkThon

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{

// Function to find the length of the longest
// subarray whose bitwise XOR is equal to K
static int longestLenXORK(int []arr,
                          int N, int K)
{

    // Stores prefix XOR
    // of the array
    int prefixXOR = 0;

    // Stores length of longest subarray
    // having bitwise XOR equal to K
    int maxLen = 0;

    // Stores index of prefix
    // XOR of the array
    Dictionary<int,
            int> mp = new Dictionary<int,
                                      int>();

    // Insert 0 into the map
    mp.Add(0, -1);

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // Update prefixXOR
        prefixXOR ^= arr[i];

        // If (prefixXOR ^ K) present
        // in the map
        if (mp.ContainsKey(prefixXOR ^ K))
        {

            // Update maxLen
            maxLen = Math.Max(maxLen,
               (i - mp[prefixXOR ^ K]));
        }

        // If prefixXOR not present
        // in the Map
        if (!mp.ContainsKey(prefixXOR))
        {

            // Insert prefixXOR
            // into the map
            mp.Add(prefixXOR, i);
        }
    }
    return maxLen;
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = {1, 2, 4, 7, 2};
    int N = arr.Length;
    int K = 1;

    Console.Write(longestLenXORK(arr, N, K));
}
}

// This code is contributed by shikhasingrajput

Javascript

<script>

// JavaScript program to implement
// the above approach

// Function to find the length of the longest
// subarray whose bitwise XOR is equal to K
function LongestLenXORK(arr, N, K)
{

    // Stores prefix XOR
    // of the array
    var prefixXOR = 0;

    // Stores length of longest subarray
    // having bitwise XOR equal to K
    var maxLen = 0;

    // Stores index of prefix
    // XOR of the array
    var mp = new Map();


    // Insert 0 into the map
    mp.set(0, -1);

    // Traverse the array
    for (var i = 0; i < N; i++) {

        // Update prefixXOR
        prefixXOR ^= arr[i];

        // If (prefixXOR ^ K) present
        // in the map
        if (mp.has(prefixXOR ^ K)) {

            // Update maxLen
            maxLen = Math.max(maxLen,
               (i - mp.get(prefixXOR ^ K)));
        }

        // If prefixXOR not present
        // in the Map
        if (!mp.has(prefixXOR)) {

            // Insert prefixXOR
            // into the map
            mp.set(prefixXOR, i);
        }
    }

    return maxLen;
}

// Driver Code
var arr = [1, 2, 4, 7, 2];
var N =  arr.length;
var K = 1;
document.write( LongestLenXORK(arr, N, K));

</script>

Output

3

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