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 3Input: 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 <bits/stdc++.h>using namespace std;// Function to find the length of the longest // subarray whose bitwise XOR is equal to Kint LongestLenXORK(int arr[], int N, int K){ //To store final answer int ans=0; //Find all subarray for(int i=0;i<N;i++){ //To store length of subarray int length=0; //To store XOR of all elements of subarray int temp=0; for(int j=i;j<N;j++){ temp=temp^arr[j]; length++; //When XOR of all elements of subarray equal to K if(temp==K){ //Update ans ans=max(ans,length); } } } return ans;}// Driver Codeint 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 approachimport 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 Kdef 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 Codeif __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 Kfunction 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 Codeconst 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 <bits/stdc++.h>using namespace std;// Function to find the length of the longest // subarray whose bitwise XOR is equal to Kint 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 Codeint 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 approachimport java.util.*;class GFG{// Function to find the length of the longest // subarray whose bitwise XOR is equal to Kstatic 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 Codepublic 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 Kdef 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 Codeif __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 approachusing System;using System.Collections.Generic;class GFG{// Function to find the length of the longest // subarray whose bitwise XOR is equal to Kstatic 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 Codepublic 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 Kfunction 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 Codevar 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)
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