Find the maximum subarray XOR in a given array (original) (raw)

Last Updated : 17 Feb, 2023

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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 value

Input: arr[] = {8, 1, 2, 12, 7, 6}
Output: 15
Explanation: The subarray {1, 2, 12} has maximum XOR value

Input: 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 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

JavaScript

`

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 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

// 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

Python3

"""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

C#

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

// 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));

`

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