Find the maximum subarray XOR in a given array - GeeksforGeeks

24 Jan, 2022
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++
Java
Python3
C#
PHP
Javascript
// 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;}
// 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
# 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 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// 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 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>
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++
Java
Python3
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 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
"""Python implementation for a Trie based solutionto find max subArray XOR""" """structure of Trie Node"""class Node:     def __init__(self, data):         self.data = data        self.left = None # left node for 0        self.right = None # right node for 1 """ 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 program to test above functions"""if __name__ == "__main__":     Arr = [8, 1, 2, 12]    n = len(Arr)    trie = Trie()    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 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
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 
Another Efficient Solution to the problem in O(n) time complexity can be achieved with the help of Kadane’s algorithm. The following algorithm is used to find the largest contagious max sum sub-array. Similarly, we can go through the array once and as we are going through it we can find the largest XOR value.
1)initialise max_now=0,max_ending=1;
2)Traverse the given array and do following for every
  array element arr[i].
      a) max_now  = bigger out of(arr[i],arr[i]^max_now)
         max_now now temprorarily contains bigger value out of
         current element and current Xor of all elements
      b) max_ending=bigger out of(max_now,max_ending) now if the max_now
         was bigger then the previous value of max_ending we update the max_ending
         else the max ending contains the biggest xor value.
C++
C
Java
Python3
C#
Javascript
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Functionint maxSubarrayXOR(int N, int arr[]){   int max_now = 0; // intitialise the 2 values  int max_e = 1;  for(int i = 0; i < N; i++){    max_e = max(arr[i], max_e ^ arr[i]);     // check whether current element/Xor is bigger    // then previous Xor value    max_now = max(max_e, max_now);  // update the maximum value  }   return max_now; // return the max value when loop is finished} // Driver codeint main (){   // Initialize the queries  int inp_arr[] = {8, 1, 2, 12};  int n = sizeof(inp_arr)/sizeof(inp_arr[0]);  cout << maxSubarrayXOR(n, inp_arr) << " is the biggest XOR value";   return 0;} // This code is contributed by Shubham Singh
// C code to implement the above approach#include <stdio.h> #define MAX(a, b) ((a) > (b) ? (a) : (b)) // Functionint maxSubarrayXOR(int N, int arr[]){         int max_now = 0; // intitialise the 2 values    int max_e = 1;    for(int i = 0; i < N; i++){        max_e = MAX(arr[i], max_e ^ arr[i]);               // check whether current element/Xor is bigger        // then previous Xor value        max_now = MAX(max_e, max_now);  // update the maximum value    }         return max_now; // return the max value when loop is finished} // Driver codeint main (){       // Initialize the queries    int inp_arr[] = {8, 1, 2, 12};    int n = sizeof(inp_arr)/sizeof(inp_arr[0]);    printf("%d is the biggest XOR value", maxSubarrayXOR(n, inp_arr));         return 0;} // This code is contributed by Shubham Singh
// Java code to implement the above approachimport java.util.*; class GFG{       // Function    public static int maxSubarrayXOR(int N, int[] arr){        int max_now = 0; // intitialise the 2 values        int max_e = 1;        for(int i = 0; i < N; i++){            max_e = Math.max(arr[i], max_e ^ arr[i]);                       // check whether current element/Xor is bigger            // then previous Xor value            max_now = Math.max(max_e, max_now);  // update the maximum value        }    return max_now; // return the max value when loop is finished    }   // Driver code  public static void main (String[] args)  {         // Initialize the queries    int[] inp_arr = {8, 1, 2, 12};    int n = inp_arr.length;    System.out.println(maxSubarrayXOR(n, inp_arr) + " is the biggest XOR value");   }} // This code is contributed by Shubham Singh
def maxSubarrayXOR(N, arr):    max_now = 0 #intitialise the 2 values    max_e = 1    for i in range(N):        max_e = max(arr[i], max_e ^ arr[i])#check whether current element/Xor is bigger                                           #then previous Xor value        max_now = max(max_e, max_now)  #update the maximum value    return max_now #return the max value when loop is finished # driver codeinp_arr = [8, 1, 2, 12]n = len(inp_arr)print(maxSubarrayXOR(n, inp_arr), "is the biggest XOR "      "value")# code contributed by# Pratyush Arora
// C# code to implement the above approachusing System;public class GFG{   // Function  public static int maxSubarrayXOR(int N, int[] arr){    int max_now = 0; // intitialise the 2 values    int max_e = 1;    for(int i = 0; i < N; i++){      max_e = Math.Max(arr[i], max_e ^ arr[i]);       // check whether current element/Xor is bigger      // then previous Xor value      max_now = Math.Max(max_e, max_now);  // update the maximum value    }    return max_now; // return the max value when loop is finished  }   // Driver code  static public void Main ()  {     // Initialize the queries    int[] inp_arr = {8, 1, 2, 12};    int n = inp_arr.Length;    Console.Write(maxSubarrayXOR(n, inp_arr) + " is the biggest XOR value");   }} // This code is contributed by Shubham Singh
<script>// Javascript o implement the above approach// Functionfunction maxSubarrayXOR(N, arr){    var max_now = 0; // intitialise the 2 values    var max_e = 1;    for(var i = 0; i < N; i++){        max_e = Math.max(arr[i], max_e ^ arr[i]);                 // check whether current element/Xor is bigger        // then previous Xor value        max_now = Math.max(max_e, max_now);  // update the maximum value    }    return max_now; // return the max value when loop is finished} // Driver code // Initialize the queriesvar inp_arr = [8, 1, 2, 12];var n = inp_arr.length;document.write(maxSubarrayXOR(n, inp_arr) +" is the biggest XOR value"); // This code is contributed by Shubham Singh</script>
12 is the biggest XOR value
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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