Given an array of integers. find the maximum XOR subarray value in given array. Expected time complexity O(n).
Examples:
Input: arr[] = {1, 2, 3, 4}
Output: 7
The subarray {3, 4} has maximum XOR value
Input: arr[] = {8, 1, 2, 12, 7, 6}
Output: 15
The subarray {1, 2, 12} has maximum XOR value
Input: arr[] = {4, 6}
Output: 6
The subarray {6} has maximum XOR value
A Simple Solution is to use two loops to find XOR of all subarrays and return the maximum.
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 functions
int 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 XOR
class 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 XOR
def 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 code
arr = [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 XOR
using 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 XOR
function 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.
?>
Output:
Max subarray XOR is 15
Time Complexity of above solution is O(n2).
An Efficient Solution can solve the above problem in O(n) time under the assumption that integers take fixed number of bits to store. The idea is to use Trie Data Structure. Below is algorithm.
1) Create an empty Trie. Every node of Trie is going to
contain two children, for 0 and 1 value of bit.
2) Initialize pre_xor = 0 and insert into the Trie.
3) Initialize result = minus infinite
4) Traverse the given array and do following for every
array element arr[i].
a) pre_xor = pre_xor ^ arr[i]
pre_xor now contains xor of elements from
arr[0] to arr[i].
b) Query the maximum xor value ending with arr[i]
from Trie.
c) Update result if the value obtained in step
4.b is more than current value of result.
How does 4.b work?
We can observe from 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.
i) 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.
ii) We need to 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, we find the entry in Trie that has maximum XOR value with current prefix. If we do XOR of such previous prefix with current prefix, we 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 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 Node
struct TrieNode
{
int value; // Only used in leaf nodes
TrieNode *arr[2];
};
// Utility function tp create a Trie node
TrieNode *newNode()
{
TrieNode *temp = new TrieNode;
temp->value = 0;
temp->arr[0] = temp->arr[1] = NULL;
return temp;
}
// Inserts pre_xor to trie with given root
void 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 functions
int 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 XOR
class 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
Output:
Max subarray XOR is 15
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 Romil Punetha. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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