# 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.

Naive Approach

The idea is to generate all subarrays and find that subarray whose bitwise XOR of all elements is equal to K and has a maximum length

Steps to Implement:

• Initialize a variable “ans” with 0 because if no such subarray exists then it will be the answer
• Run two for loops from 0 to N-1 to generate all subarray
• For each subarray find the XOR of all elements and its length
• If any XOR value got equal to K then update “ans” as the maximum of “ans” and the length of that subarray
• In the last print/return value in ans

Code-

## C++

 `// C++ program to implement` `// the above approach`   `#include ` `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)` `{` `    ``//To store final answer` `    ``int` `ans=0;` `    `  `    ``//Find all subarray` `    ``for``(``int` `i=0;i

## 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)` `    ``{` `        ``// To store the final answer` `        ``int` `ans = ``0``;`   `        ``// Find all subarrays` `        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``// To store the length of the subarray` `            ``int` `length = ``0``;` `            ``// To store XOR of all elements of the subarray` `            ``int` `temp = ``0``;` `            ``for` `(``int` `j = i; j < N; j++) {` `                ``temp = temp ^ arr[j];` `                ``length++;`   `                ``// When XOR of all elements of subarray is` `                ``// equal to K` `                ``if` `(temp == K) {` `                    ``// Update ans` `                    ``ans = Math.max(ans, length);` `                ``}` `            ``}` `        ``}`   `        ``return` `ans;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int``[] arr = { ``1``, ``2``, ``4``, ``7``, ``2` `};` `        ``int` `N = arr.length;` `        ``int` `K = ``1``;` `        ``System.out.println(LongestLenXORK(arr, N, K));` `    ``}` `}`

## Python3

 `# Python 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):` `    ``# To store final answer` `    ``ans ``=` `0`   `    ``# Find all subarray` `    ``for` `i ``in` `range``(N):` `        ``# To store length of subarray` `        ``length ``=` `0` `        ``# To store XOR of all elements of subarray` `        ``temp ``=` `0` `        ``for` `j ``in` `range``(i, N):` `            ``temp ^``=` `arr[j]` `            ``length ``+``=` `1`   `            ``# When XOR of all elements of subarray equal to K` `            ``if` `temp ``=``=` `K:` `                ``# Update ans` `                ``ans ``=` `max``(ans, length)`   `    ``return` `ans`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``1``, ``2``, ``4``, ``7``, ``2``]` `    ``N ``=` `len``(arr)` `    ``K ``=` `1` `    ``print``(LongestLenXORK(arr, N, K))`

## C#

 `using` `System;`   `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)` `    ``{` `        ``// To store the final answer` `        ``int` `ans = 0;`   `        ``// Find all subarrays` `        ``for` `(``int` `i = 0; i < N; i++)` `        ``{` `            ``// To store the length of the subarray` `            ``int` `length = 0;` `            `  `            ``// To store the XOR of all elements of the subarray` `            ``int` `temp = 0;`   `            ``for` `(``int` `j = i; j < N; j++)` `            ``{` `                ``temp ^= arr[j];` `                ``length++;`   `                ``// When XOR of all elements of the subarray is equal to K` `                ``if` `(temp == K)` `                ``{` `                    ``// Update ans` `                    ``ans = Math.Max(ans, length);` `                ``}` `            ``}` `        ``}`   `        ``return` `ans;` `    ``}`   `    ``// Driver Code` `    ``static` `void` `Main(``string``[] args)` `    ``{` `        ``int``[] arr = { 1, 2, 4, 7, 2 };` `        ``int` `N = arr.Length;` `        ``int` `K = 1;` `        ``Console.WriteLine(LongestLenXORK(arr, N, K));` `    ``}` `}`   `// This code is contributed by Dwaipayan Bandyopadhyay`

## Javascript

 `// 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) {` ` ``//To store final answer` `    ``let ans = 0;` `//Find all subarray` `    ``for` `(let i = 0; i < N; i++) {` `        ``let length = 0;` `        ``//To store XOR of all elements of subarray` `        ``let temp = 0;` `        ``for` `(let j = i; j < N; j++) {` `            ``temp ^= arr[j];` `            ``length += 1;` `//When XOR of all elements of subarray equal to K` `            ``if` `(temp === K) {` `            ``//Update ans` `                ``ans = Math.max(ans, length);` `            ``}` `        ``}` `    ``}` `    ``return` `ans;` `}`   `// Driver Code` `const arr = [1, 2, 4, 7, 2];` `const N = arr.length;` `const K = 1;` `console.log(LongestLenXORK(arr, N, K));`

Output-

`3`

Time Complexity: O(N2), because of two nested loops from 0 to N-1
Auxiliary Space: O(1), because no extra space has been used

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

a1 ^ a2 ^ a3 ^ ….. ^ an = K

=> a2 ^ a3 ^ ….. ^ an ^ 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 ith 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 ith 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 ` `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 mp = ``new` `HashMap();` `                                      `  `    ``// 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

 ``

Output

```3

```

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

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