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:
// 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;
} |
// 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 |
# 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. |
// 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 // 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. ?> |
<script> // A simple Javascript program to find // max subarray XOR function 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 code let 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> |
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++ 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 to 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 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 |
"""Python implementation for a Trie based solution to find max subArray XOR""" # Structure of Trie Node class 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 Trie class 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 code if __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 |
using System;
// C# program for a Trie based O(n) solution to // find max subarray XOR public 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 program for a Trie based O(n) solution to find max subarray XOR // Assumed int size const INT_SIZE = 32; // A Trie Node class TrieNode { constructor() {
this .value = 0; // Only used in leaf nodes
this .arr = [ null , null ];
}
} let root; // Inserts pre_xor to trie with given root function 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 functions let arr = [8, 1, 2, 12]; let n = arr.length; console.log( "Max subarray XOR is " + maxSubarrayXOR(arr, n));
|
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