Given an array of integers. The task is to find the maximum subarray XOR value in the given array.
Examples:
Input: arr[] = {1, 2, 3, 4}
Output: 7
Explanation: The subarray {3, 4} has maximum XOR valueInput: arr[] = {8, 1, 2, 12, 7, 6}
Output: 15
Explanation: The subarray {1, 2, 12} has maximum XOR valueInput: arr[] = {4, 6}
Output: 6
Explanation: The subarray {6} has maximum XOR value
Naive Approach: Below is the idea to solve the problem:
Create all possible subarrays and calculate the XOR of the subarrays. The maximum among them will be the required answer.
Follow the steps mentioned below to implement the idea:
- Iterate from i = 0 to N-1:
- Initialize a variable (say curr_xor = 0) to store the XOR value of subarrays starting from i
- Run a nested loop from j = i to N-1:
- The value j determines the ending point for the current subarray starting from i.
- Update curr_xor by performing XOR of curr_xor with arr[j].
- If the value is greater than the maximum then update the maximum value also.
- The maximum value is the required answer.
Below is the Implementation of the above approach:
C++
// A simple C++ program to find max subarray XOR#include<bits/stdc++.h>using namespace std;int maxSubarrayXOR(int arr[], int n){ int ans = INT_MIN; // Initialize result // Pick starting points of subarrays for (int i=0; i<n; i++) { int curr_xor = 0; // to store xor of current subarray // Pick ending points of subarrays starting with i for (int j=i; j<n; j++) { curr_xor = curr_xor ^ arr[j]; ans = max(ans, curr_xor); } } return ans;}// Driver program to test above functionsint main(){ int arr[] = {8, 1, 2, 12}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Max subarray XOR is " << maxSubarrayXOR(arr, n); return 0;} |
Java
// A simple Java program to find max subarray XORclass GFG { static int maxSubarrayXOR(int arr[], int n) { int ans = Integer.MIN_VALUE; // Initialize result // Pick starting points of subarrays for (int i=0; i<n; i++) { // to store xor of current subarray int curr_xor = 0; // Pick ending points of subarrays starting with i for (int j=i; j<n; j++) { curr_xor = curr_xor ^ arr[j]; ans = Math.max(ans, curr_xor); } } return ans; } // Driver program to test above functions public static void main(String args[]) { int arr[] = {8, 1, 2, 12}; int n = arr.length; System.out.println("Max subarray XOR is " + maxSubarrayXOR(arr, n)); }}//This code is contributed by Sumit Ghosh |
Python3
# A simple Python program# to find max subarray XORdef maxSubarrayXOR(arr,n): ans = -2147483648 #Initialize result # Pick starting points of subarrays for i in range(n): # to store xor of current subarray curr_xor = 0 # Pick ending points of # subarrays starting with i for j in range(i,n): curr_xor = curr_xor ^ arr[j] ans = max(ans, curr_xor) return ans# Driver codearr = [8, 1, 2, 12]n = len(arr)print("Max subarray XOR is ", maxSubarrayXOR(arr, n))# This code is contributed# by Anant Agarwal. |
C#
// A simple C# program to find// max subarray XORusing System;class GFG{ // Function to find max subarray static int maxSubarrayXOR(int []arr, int n) { int ans = int.MinValue; // Initialize result // Pick starting points of subarrays for (int i = 0; i < n; i++) { // to store xor of current subarray int curr_xor = 0; // Pick ending points of // subarrays starting with i for (int j = i; j < n; j++) { curr_xor = curr_xor ^ arr[j]; ans = Math.Max(ans, curr_xor); } } return ans; } // Driver code public static void Main() { int []arr = {8, 1, 2, 12}; int n = arr.Length; Console.WriteLine("Max subarray XOR is " + maxSubarrayXOR(arr, n)); }}// This code is contributed by Sam007. |
PHP
<?php// A simple PHP program to// find max subarray XORfunction maxSubarrayXOR($arr, $n){ // Initialize result $ans = PHP_INT_MIN; // Pick starting points // of subarrays for ($i = 0; $i < $n; $i++) { // to store xor of // current subarray $curr_xor = 0; // Pick ending points of // subarrays starting with i for ($j = $i; $j < $n; $j++) { $curr_xor = $curr_xor ^ $arr[$j]; $ans = max($ans, $curr_xor); } } return $ans;} // Driver Code $arr = array(8, 1, 2, 12); $n = count($arr); echo "Max subarray XOR is " , maxSubarrayXOR($arr, $n); // This code is contributed by anuj_67.?> |
Javascript
<script>// A simple Javascript program to find// max subarray XORfunction maxSubarrayXOR(arr, n){ // Initialize result let ans = Number.MIN_VALUE; // Pick starting points of subarrays for(let i = 0; i < n; i++) { // To store xor of current subarray let curr_xor = 0; // Pick ending points of subarrays // starting with i for(let j = i; j < n; j++) { curr_xor = curr_xor ^ arr[j]; ans = Math.max(ans, curr_xor); } } return ans;}// Driver codelet arr = [ 8, 1, 2, 12 ];let n = arr.length;document.write("Max subarray XOR is " + maxSubarrayXOR(arr, n)); // This code is contributed by divyesh072019</script> |
Output
Max subarray XOR is 15
Time Complexity: O(N2).
Auxiliary Space: O(1)
Find the maximum subarray XOR in a given array using trie Data Structure.
Maximize the xor subarray by using trie data structure to find the binary inverse of current prefix xor inorder to set the left most unset bits and maximize the value.
Follow the below steps to Implement the idea:
- Create an empty Trie. Every node of Trie is going to contain two children, for 0 and 1 values of a bit.
- Initialize pre_xor = 0 and insert into the Trie, Initialize result = INT_MIN
- Traverse the given array and do the following for every array element arr[i].
- pre_xor = pre_xor ^ arr[i], pre_xor now contains xor of elements from arr[0] to arr[i].
- Query the maximum xor value ending with arr[i] from Trie.
- Update the result if the value obtained above is more than the current value of the result.
Illustration:
It can be observed from the above algorithm that we build a Trie that contains XOR of all prefixes of given array. To find the maximum XOR subarray ending with arr[i], there may be two cases.
- The prefix itself has the maximum XOR value ending with arr[i]. For example if i=2 in {8, 2, 1, 12}, then the maximum subarray xor ending with arr[2] is the whole prefix.
- Remove some prefix (ending at index from 0 to i-1). For example if i=3 in {8, 2, 1, 12}, then the maximum subarray xor ending with arr[3] starts with arr[1] and we need to remove arr[0].
- To find the prefix to be removed, find the entry in Trie that has maximum XOR value with current prefix. If we do XOR of such previous prefix with current prefix, get the maximum XOR value ending with arr[i].
- If there is no prefix to be removed (case i), then we return 0 (that’s why we inserted 0 in Trie).
Below is the implementation of the above idea :
C++
// C++ program for a Trie based O(n) solution to find max// subarray XOR#include<bits/stdc++.h>using namespace std;// Assumed int size#define INT_SIZE 32// A Trie Nodestruct TrieNode{ int value; // Only used in leaf nodes TrieNode *arr[2];};// Utility function to create a Trie nodeTrieNode *newNode(){ TrieNode *temp = new TrieNode; temp->value = 0; temp->arr[0] = temp->arr[1] = NULL; return temp;}// Inserts pre_xor to trie with given rootvoid insert(TrieNode *root, int pre_xor){ TrieNode *temp = root; // Start from the msb, insert all bits of // pre_xor into Trie for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix bool val = pre_xor & (1<<i); // Create a new node if needed if (temp->arr[val] == NULL) temp->arr[val] = newNode(); temp = temp->arr[val]; } // Store value at leaf node temp->value = pre_xor;}// Finds the maximum XOR ending with last number in// prefix XOR 'pre_xor' and returns the XOR of this maximum// with pre_xor which is maximum XOR ending with last element// of pre_xor.int query(TrieNode *root, int pre_xor){ TrieNode *temp = root; for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix bool val = pre_xor & (1<<i); // Traverse Trie, first look for a // prefix that has opposite bit if (temp->arr[1-val]!=NULL) temp = temp->arr[1-val]; // If there is no prefix with opposite // bit, then look for same bit. else if (temp->arr[val] != NULL) temp = temp->arr[val]; } return pre_xor^(temp->value);}// Returns maximum XOR value of a subarray in arr[0..n-1]int maxSubarrayXOR(int arr[], int n){ // Create a Trie and insert 0 into it TrieNode *root = newNode(); insert(root, 0); // Initialize answer and xor of current prefix int result = INT_MIN, pre_xor =0; // Traverse all input array element for (int i=0; i<n; i++) { // update current prefix xor and insert it into Trie pre_xor = pre_xor^arr[i]; insert(root, pre_xor); // Query for current prefix xor in Trie and update // result if required result = max(result, query(root, pre_xor)); } return result;}// Driver program to test above functionsint main(){ int arr[] = {8, 1, 2, 12}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Max subarray XOR is " << maxSubarrayXOR(arr, n); return 0;} |
Java
// Java program for a Trie based O(n) solution to// find max subarray XORclass GFG{ // Assumed int size static final int INT_SIZE = 32; // A Trie Node static class TrieNode { int value; // Only used in leaf nodes TrieNode[] arr = new TrieNode[2]; public TrieNode() { value = 0; arr[0] = null; arr[1] = null; } } static TrieNode root; // Inserts pre_xor to trie with given root static void insert(int pre_xor) { TrieNode temp = root; // Start from the msb, insert all bits of // pre_xor into Trie for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix int val = (pre_xor & (1<<i)) >=1 ? 1 : 0; // Create a new node if needed if (temp.arr[val] == null) temp.arr[val] = new TrieNode(); temp = temp.arr[val]; } // Store value at leaf node temp.value = pre_xor; } // Finds the maximum XOR ending with last number in // prefix XOR 'pre_xor' and returns the XOR of this // maximum with pre_xor which is maximum XOR ending // with last element of pre_xor. static int query(int pre_xor) { TrieNode temp = root; for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix int val = (pre_xor & (1<<i)) >= 1 ? 1 : 0; // Traverse Trie, first look for a // prefix that has opposite bit if (temp.arr[1-val] != null) temp = temp.arr[1-val]; // If there is no prefix with opposite // bit, then look for same bit. else if (temp.arr[val] != null) temp = temp.arr[val]; } return pre_xor^(temp.value); } // Returns maximum XOR value of a subarray in // arr[0..n-1] static int maxSubarrayXOR(int arr[], int n) { // Create a Trie and insert 0 into it root = new TrieNode(); insert(0); // Initialize answer and xor of current prefix int result = Integer.MIN_VALUE; int pre_xor =0; // Traverse all input array element for (int i=0; i<n; i++) { // update current prefix xor and insert it // into Trie pre_xor = pre_xor^arr[i]; insert(pre_xor); // Query for current prefix xor in Trie and // update result if required result = Math.max(result, query(pre_xor)); } return result; } // Driver program to test above functions public static void main(String args[]) { int arr[] = {8, 1, 2, 12}; int n = arr.length; System.out.println("Max subarray XOR is " + maxSubarrayXOR(arr, n)); }}// This code is contributed by Sumit Ghosh |
Python3
"""Python implementation for a Trie based solutionto find max subArray XOR"""# Structure of Trie Nodeclass Node: def __init__(self, data): self.data = data # left node for 0 self.left = None # right node for 1 self.right = None# Class for implementing Trieclass Trie: def __init__(self): self.root = Node(0) # Insert pre_xor to trie with given root def insert(self, pre_xor): self.temp = self.root # Start from msb, insert all bits of pre_xor # into the Trie for i in range(31, -1, -1): # Find current bit in prefix sum val = pre_xor & (1<<i) if val : # Create new node if needed if not self.temp.right: self.temp.right = Node(0) self.temp = self.temp.right if not val: # Create new node if needed if not self.temp.left: self.temp.left = Node(0) self.temp = self.temp.left # Store value at leaf node self.temp.data = pre_xor # Find the maximum xor ending with last number # in prefix XOR and return the XOR of this def query(self, xor): self.temp = self.root for i in range(31, -1, -1): # Find the current bit in prefix xor val = xor & (1<<i) # Traverse the trie, first look for opposite bit # and then look for same bit if val: if self.temp.left: self.temp = self.temp.left elif self.temp.right: self.temp = self.temp.right else: if self.temp.right: self.temp = self.temp.right elif self.temp.left: self.temp = self.temp.left return xor ^ self.temp.data # Returns maximum XOR value of subarray def maxSubArrayXOR(self, n, Arr): # Insert 0 in the trie self.insert(0) # Initialize result and pre_xor result = -float('inf') pre_xor = 0 # Traverse all input array element for i in range(n): # Update current prefix xor and # insert it into Trie pre_xor = pre_xor ^ Arr[i] self.insert(pre_xor) # Query for current prefix xor # in Trie and update result result = max(result, self.query(pre_xor)) return result# Driver codeif __name__ == "__main__": Arr = [8, 1, 2, 12] n = len(Arr) trie = Trie() print("Max subarray XOR is", end = ' ') print(trie.maxSubArrayXOR(n, Arr))# This code is contributed by chaudhary_19 |
C#
using System;// C# program for a Trie based O(n) solution to // find max subarray XORpublic class GFG{ // Assumed int size public const int INT_SIZE = 32; // A Trie Node public class TrieNode { public int value; // Only used in leaf nodes public TrieNode[] arr = new TrieNode[2]; public TrieNode() { value = 0; arr[0] = null; arr[1] = null; } } public static TrieNode root; // Inserts pre_xor to trie with given root public static void insert(int pre_xor) { TrieNode temp = root; // Start from the msb, insert all bits of // pre_xor into Trie for (int i = INT_SIZE-1; i >= 0; i--) { // Find current bit in given prefix int val = (pre_xor & (1 << i)) >= 1 ? 1 : 0; // Create a new node if needed if (temp.arr[val] == null) { temp.arr[val] = new TrieNode(); } temp = temp.arr[val]; } // Store value at leaf node temp.value = pre_xor; } // Finds the maximum XOR ending with last number in // prefix XOR 'pre_xor' and returns the XOR of this // maximum with pre_xor which is maximum XOR ending // with last element of pre_xor. public static int query(int pre_xor) { TrieNode temp = root; for (int i = INT_SIZE-1; i >= 0; i--) { // Find current bit in given prefix int val = (pre_xor & (1 << i)) >= 1 ? 1 : 0; // Traverse Trie, first look for a // prefix that has opposite bit if (temp.arr[1 - val] != null) { temp = temp.arr[1 - val]; } // If there is no prefix with opposite // bit, then look for same bit. else if (temp.arr[val] != null) { temp = temp.arr[val]; } } return pre_xor ^ (temp.value); } // Returns maximum XOR value of a subarray in // arr[0..n-1] public static int maxSubarrayXOR(int[] arr, int n) { // Create a Trie and insert 0 into it root = new TrieNode(); insert(0); // Initialize answer and xor of current prefix int result = int.MinValue; int pre_xor = 0; // Traverse all input array element for (int i = 0; i < n; i++) { // update current prefix xor and insert it // into Trie pre_xor = pre_xor ^ arr[i]; insert(pre_xor); // Query for current prefix xor in Trie and // update result if required result = Math.Max(result, query(pre_xor)); } return result; } // Driver program to test above functions public static void Main(string[] args) { int[] arr = new int[] {8, 1, 2, 12}; int n = arr.Length; Console.WriteLine("Max subarray XOR is " + maxSubarrayXOR(arr, n)); }} // This code is contributed by Shrikant13 |
Javascript
// JavaScript program for a Trie based O(n) solution to find max subarray XOR// Assumed int sizeconst INT_SIZE = 32;// A Trie Nodeclass TrieNode { constructor() { this.value = 0; // Only used in leaf nodes this.arr = [null, null]; }}let root;// Inserts pre_xor to trie with given rootfunction insert(pre_xor) { let temp = root; // Start from the msb, insert all bits of // pre_xor into Trie for (let i = INT_SIZE - 1; i >= 0; i--) { // Find current bit in given prefix let val = (pre_xor & (1 << i)) >= 1 ? 1 : 0; // Create a new node if needed if (temp.arr[val] === null) { temp.arr[val] = new TrieNode(); } temp = temp.arr[val]; } // Store value at leaf node temp.value = pre_xor;}// Finds the maximum XOR ending with last number in// prefix XOR 'pre_xor' and returns the XOR of this// maximum with pre_xor which is maximum XOR ending// with last element of pre_xor.function query(pre_xor) { let temp = root; for (let i = INT_SIZE - 1; i >= 0; i--) { // Find current bit in given prefix let val = (pre_xor & (1 << i)) >= 1 ? 1 : 0; // Traverse Trie, first look for a // prefix that has opposite bit if (temp.arr[1 - val] !== null) { temp = temp.arr[1 - val]; } // If there is no prefix with opposite // bit, then look for same bit. else if (temp.arr[val] !== null) { temp = temp.arr[val]; } } return pre_xor ^ temp.value;}// Returns maximum XOR value of a subarray in arr[0..n-1]function maxSubarrayXOR(arr, n) { // Create a Trie and insert 0 into it root = new TrieNode(); insert(0); // Initialize answer and xor of current prefix let result = Number.MIN_SAFE_INTEGER; let pre_xor = 0; // Traverse all input array element for (let i = 0; i < n; i++) { // update current prefix xor and insert it // into Trie pre_xor ^= arr[i]; insert(pre_xor); // Query for current prefix xor in Trie and // update result if required result = Math.max(result, query(pre_xor)); } return result;}// Driver program to test above functionslet arr = [8, 1, 2, 12];let n = arr.length;console.log("Max subarray XOR is " + maxSubarrayXOR(arr, n)); |
Output
Max subarray XOR is 15
Time Complexity: O(N).
Auxiliary Space: O(N)
Exercise: Extend the above solution so that it also prints starting and ending indexes of subarray with maximum value (Hint: we can add one more field to Trie node to achieve this
This article is contributed by Aarti_Rathi and Romil Punetha. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
