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 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.
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 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 |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(N)
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