Length of the longest subarray whose Bitwise XOR is K
Last Updated :
26 Sep, 2023
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 <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)
{
//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 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 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));
}
}
# 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))
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 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 <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 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 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# 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
<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>
Time Complexity: O(N)
Auxiliary Space: O(N)
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