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Count the number of subarrays having a given XOR

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  • Difficulty Level : Hard
  • Last Updated : 19 Jul, 2022
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Given an array of integers arr[] and a number m, count the number of subarrays having XOR of their elements as m.
Examples: 

Input : arr[] = {4, 2, 2, 6, 4}, m = 6
Output : 4
Explanation : The subarrays having XOR of 
              their elements as 6 are {4, 2}, 
              {4, 2, 2, 6, 4}, {2, 2, 6},
               and {6}

Input : arr[] = {5, 6, 7, 8, 9}, m = 5
Output : 2
Explanation : The subarrays having XOR of
              their elements as 5 are {5}
              and {5, 6, 7, 8, 9}

A Simple Solution is to use two loops to go through all possible subarrays of arr[] and count the number of subarrays having XOR of their elements as m. 

Implementation:

C++




// A simple C++ Program to count all subarrays having
// XOR of elements as given value m
#include <bits/stdc++.h>
using namespace std;
 
// Simple function that returns count of subarrays
// of arr with XOR value equals to m
long long subarrayXor(int arr[], int n, int m)
{
    long long ans = 0; // Initialize ans
 
    // Pick starting point i of subarrays
    for (int i = 0; i < n; i++) {
        int xorSum = 0; // Store XOR of current subarray
 
        // Pick ending point j of subarray for each i
        for (int j = i; j < n; j++) {
            // calculate xorSum
            xorSum = xorSum ^ arr[j];
 
            // If xorSum is equal to given value,
            // increase ans by 1.
            if (xorSum == m)
                ans++;
        }
    }
    return ans;
}
 
// Driver program to test above function
int main()
{
    int arr[] = { 4, 2, 2, 6, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int m = 6;
 
    cout << "Number of subarrays having given XOR is "
         << subarrayXor(arr, n, m);
    return 0;
}


Java




// A simple Java Program to count all
// subarrays having XOR of elements
// as given value m
public class GFG {
 
    // Simple function that returns
    // count of subarrays of arr with
    // XOR value equals to m
    static long subarrayXor(int arr[],
                             int n, int m)
    {
         
        // Initialize ans
        long ans = 0;
 
        // Pick starting point i of
        // subarrays
        for (int i = 0; i < n; i++)
        {
             
            // Store XOR of current
            // subarray
            int xorSum = 0;
 
            // Pick ending point j of
            // subarray for each i
            for (int j = i; j < n; j++)
            {
                 
                // calculate xorSum
                xorSum = xorSum ^ arr[j];
 
                // If xorSum is equal to
                // given value, increase
                // ans by 1.
                if (xorSum == m)
                    ans++;
            }
        }
         
        return ans;
    }
 
    // Driver code
    public static void main(String args[])
    {
 
        int[] arr = { 4, 2, 2, 6, 4 };
        int n = arr.length;
        int m = 6;
 
        System.out.println("Number of subarrays"
                       + " having given XOR is "
                       + subarrayXor(arr, n, m));
    }
}
 
// This code is contributed by Sam007.


Python3




     
# A simple Python3 Program to count all subarrays having
# XOR of elements as given value m
  
# Simple function that returns count of subarrays
# of arr with XOR value equals to m
def subarrayXor(arr, n, m):
    ans = 0 # Initialize ans
  
    # Pick starting point i of subarrays
    for i in range(0,n):
         
        xorSum = 0 # Store XOR of current subarray
  
        # Pick ending point j of subarray for each i
        for in range(i,n):
            # calculate xorSum
            xorSum = xorSum ^ arr[j]
  
            # If xorSum is equal to given value,
            # increase ans by 1.
            if (xorSum == m):
                ans+=1
    return ans
  
# Driver program to test above function
def main():
    arr = [ 4, 2, 2, 6, 4 ]
    n = len(arr)
    m = 6
  
    print("Number of subarrays having given XOR is "
         , subarrayXor(arr, n, m))
 
if __name__ == '__main__':
    main()
     
#this code contributed by 29AjayKumar


C#




// A simple C# Program to count all
// subarrays having XOR of elements
// as given value m
using System;
 
class GFG {
     
    // Simple function that returns
    // count of subarrays of arr
    // with XOR value equals to m
    static long subarrayXor(int[] arr,
                            int n, int m)
    {
         
        // Initialize ans
        long ans = 0;
 
        // Pick starting point i of
        // subarrays
        for (int i = 0; i < n; i++)
        {
             
            // Store XOR of current
            // subarray
            int xorSum = 0;
 
            // Pick ending point j of
            // subarray for each i
            for (int j = i; j < n; j++)
            {
                 
                // calculate xorSum
                xorSum = xorSum ^ arr[j];
 
                // If xorSum is equal to
                // given value, increase
                // ans by 1.
                if (xorSum == m)
                    ans++;
            }
        }
         
        return ans;
    }
 
    // Driver Program
    public static void Main()
    {
        int[] arr = { 4, 2, 2, 6, 4 };
        int n = arr.Length;
        int m = 6;
 
        Console.Write("Number of subarrays"
                  + " having given XOR is "
                  + subarrayXor(arr, n, m));
    }
}
 
// This code is contributed by Sam007.


PHP




<?php
// A simple PHP Program to
// count all subarrays having
// XOR of elements as given value m
 
// Simple function that returns
// count of subarrays of arr
// with XOR value equals to m
function subarrayXor($arr, $n,$m)
{
     
    // Initialize ans
    $ans = 0;
 
    // Pick starting point
    // i of subarrays
    for ($i = 0; $i < $n; $i++)
    {
         
        // Store XOR of
        // current subarray
        $xorSum = 0;
 
        // Pick ending point j of
        // subarray for each i
        for ($j = $i; $j < $n; $j++)
        {
            // calculate xorSum
            $xorSum = $xorSum ^ $arr[$j];
 
            // If xorSum is equal
            // to given value,
            // increase ans by 1.
            if ($xorSum == $m)
                $ans++;
        }
    }
    return $ans;
}
 
    // Driver Code
    $arr = array(4, 2, 2, 6, 4);
    $n = count($arr);
    $m = 6;
 
    echo "Number of subarrays having given XOR is "
         , subarrayXor($arr, $n, $m);
          
// This code is contributed by anuj_67.
?>


Javascript




<script>
 
// A simple Javascript Program to
// count all subarrays having
// XOR of elements as given value m
 
// Simple function that
// returns count of subarrays
// of arr with XOR value equals to m
function subarrayXor(arr, n, m)
{
    let ans = 0; // Initialize ans
 
    // Pick starting point i of subarrays
    for (let i = 0; i < n; i++)
    {
    // Store XOR of current subarray
        let xorSum = 0;
 
        // Pick ending point j of
        // subarray for each i
        for (let j = i; j < n; j++) {
            // calculate xorSum
            xorSum = xorSum ^ arr[j];
 
            // If xorSum is equal to given value,
            // increase ans by 1.
            if (xorSum == m)
                ans++;
        }
    }
    return ans;
}
 
// Driver program to test above function
    let arr = [ 4, 2, 2, 6, 4 ];
    let n = arr.length;
    let m = 6;
 
    document.write(
    "Number of subarrays having given XOR is "
         + subarrayXor(arr, n, m)
         );
 
</script>


Output

Number of subarrays having given XOR is 4

Time Complexity: O(n2)

Auxiliary Space: O(1)

Efficient Approach:

An Efficient Solution solves the above problem in O(n) time. Let us call the XOR of all elements in the range [i+1, j] as A, in the range [0, i] as B, and in the range [0, j] as C. If we do XOR of B with C, the overlapping elements in [0, i] from B and C zero out, and we get XOR of all elements in the range [i+1, j], i.e. A. Since A = B XOR C, we have B = A XOR C. Now, if we know the value of C and we take the value of A as m, we get the count of A as the count of all B satisfying this relation. Essentially, we get the count of all subarrays having XOR-sum m for each C. As we take the sum of this count overall C, we get our answer.

1) Initialize ans as 0.
2) Compute xorArr, the prefix xor-sum array.
3) Create a map mp in which we store count of 
   all prefixes with XOR as a particular value. 
4) Traverse xorArr and for each element in xorArr
   (A) If m^xorArr[i] XOR exists in map, then 
       there is another previous prefix with 
       same XOR, i.e., there is a subarray ending
       at i with XOR equal to m. We add count of
       all such subarrays to result. 
   (B) If xorArr[i] is equal to m, increment ans by 1.
   (C) Increment count of elements having XOR-sum 
       xorArr[i] in map by 1.
5) Return ans.

Implementation:

C++




// C++ Program to count all subarrays having
// XOR of elements as given value m with
// O(n) time complexity.
#include <bits/stdc++.h>
using namespace std;
 
// Returns count of subarrays of arr with XOR
// value equals to m
long long subarrayXor(int arr[], int n, int m)
{
    long long ans = 0; // Initialize answer to be returned
 
    // Create a prefix xor-sum array such that
    // xorArr[i] has value equal to XOR
    // of all elements in arr[0 ..... i]
    int* xorArr = new int[n];
 
    // Create map that stores number of prefix array
    // elements corresponding to a XOR value
    unordered_map<int, int> mp;
 
    // Initialize first element of prefix array
    xorArr[0] = arr[0];
 
    // Computing the prefix array.
    for (int i = 1; i < n; i++)
        xorArr[i] = xorArr[i - 1] ^ arr[i];
 
    // Calculate the answer
    for (int i = 0; i < n; i++) {
       
        // Find XOR of current prefix with m.
        int tmp = m ^ xorArr[i];
 
        // If above XOR exists in map, then there
        // is another previous prefix with same
        // XOR, i.e., there is a subarray ending
        // at i with XOR equal to m.
        ans = ans + ((long long)mp[tmp]);
 
        // If this subarray has XOR equal to m itself.
        if (xorArr[i] == m)
            ans++;
 
        // Add the XOR of this subarray to the map
        mp[xorArr[i]]++;
    }
 
    // Return total count of subarrays having XOR of
    // elements as given value m
    return ans;
}
 
// Driver program to test above function
int main()
{
    int arr[] = { 4, 2, 2, 6, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int m = 6;
 
    cout << "Number of subarrays having given XOR is "
         << subarrayXor(arr, n, m);
    return 0;
}


Java




// Java Program to count all subarrays having
// XOR of elements as given value m with
// O(n) time complexity.
import java.util.*;
 
class GFG {
 
    // Returns count of subarrays of arr with XOR
    // value equals to m
    static long subarrayXor(int arr[], int n, int m)
    {
        long ans = 0; // Initialize answer to be returned
 
        // Create a prefix xor-sum array such that
        // xorArr[i] has value equal to XOR
        // of all elements in arr[0 ..... i]
        int[] xorArr = new int[n];
 
        // Create map that stores number of prefix array
        // elements corresponding to a XOR value
        HashMap<Integer, Integer> mp
            = new HashMap<Integer, Integer>();
 
        // Initialize first element of prefix array
        xorArr[0] = arr[0];
 
        // Computing the prefix array.
        for (int i = 1; i < n; i++)
            xorArr[i] = xorArr[i - 1] ^ arr[i];
 
        // Calculate the answer
        for (int i = 0; i < n; i++) {
            // Find XOR of current prefix with m.
            int tmp = m ^ xorArr[i];
 
            // If above XOR exists in map, then there
            // is another previous prefix with same
            // XOR, i.e., there is a subarray ending
            // at i with XOR equal to m.
            ans = ans
                  + (mp.containsKey(tmp) == false
                         ? 0
                         : ((long)mp.get(tmp)));
 
            // If this subarray has XOR equal to m itself.
            if (xorArr[i] == m)
                ans++;
 
            // Add the XOR of this subarray to the map
            if (mp.containsKey(xorArr[i]))
                mp.put(xorArr[i], mp.get(xorArr[i]) + 1);
            else
                mp.put(xorArr[i], 1);
        }
 
        // Return total count of subarrays having XOR of
        // elements as given value m
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 4, 2, 2, 6, 4 };
        int n = arr.length;
        int m = 6;
 
        System.out.print(
            "Number of subarrays having given XOR is "
            + subarrayXor(arr, n, m));
    }
}
 
// This code is contributed by PrinciRaj1992


Python3




# Python3 Program to count all subarrays
# having XOR of elements as given value m
# with O(n) time complexity.
 
# Returns count of subarrays of arr
# with XOR value equals to m
def subarrayXor(arr, n, m):
 
    ans = 0 # Initialize answer to be returned
 
    # Create a prefix xor-sum array such that
    # xorArr[i] has value equal to XOR
    # of all elements in arr[0 ..... i]
    xorArr =[0 for _ in range(n)]
 
    # Create map that stores number of prefix array
    # elements corresponding to a XOR value
    mp = dict()
 
    # Initialize first element
    # of prefix array
    xorArr[0] = arr[0]
 
    # Computing the prefix array.
    for i in range(1, n):
        xorArr[i] = xorArr[i - 1] ^ arr[i]
 
    # Calculate the answer
    for i in range(n):
         
        # Find XOR of current prefix with m.
        tmp = m ^ xorArr[i]
 
        # If above XOR exists in map, then there
        # is another previous prefix with same
        # XOR, i.e., there is a subarray ending
        # at i with XOR equal to m.
        if tmp in mp.keys():
            ans = ans + (mp[tmp])
 
        # If this subarray has XOR
        # equal to m itself.
        if (xorArr[i] == m):
            ans += 1
 
        # Add the XOR of this subarray to the map
        mp[xorArr[i]] = mp.get(xorArr[i], 0) + 1
 
    # Return total count of subarrays having
    # XOR of elements as given value m
    return ans
 
# Driver Code
arr = [4, 2, 2, 6, 4]
n = len(arr)
m = 6
 
print("Number of subarrays having given XOR is",
                        subarrayXor(arr, n, m))
 
# This code is contributed by mohit kumar


C#




// C# Program to count all subarrays having
// XOR of elements as given value m with
// O(n) time complexity.
using System;
using System.Collections.Generic;
 
class GFG {
 
    // Returns count of subarrays of arr with XOR
    // value equals to m
    static long subarrayXor(int[] arr, int n, int m)
    {
        long ans = 0; // Initialize answer to be returned
 
        // Create a prefix xor-sum array such that
        // xorArr[i] has value equal to XOR
        // of all elements in arr[0 ..... i]
        int[] xorArr = new int[n];
 
        // Create map that stores number of prefix array
        // elements corresponding to a XOR value
        Dictionary<int, int> mp
            = new Dictionary<int, int>();
 
        // Initialize first element of prefix array
        xorArr[0] = arr[0];
 
        // Computing the prefix array.
        for (int i = 1; i < n; i++)
            xorArr[i] = xorArr[i - 1] ^ arr[i];
 
        // Calculate the answer
        for (int i = 0; i < n; i++) {
            // Find XOR of current prefix with m.
            int tmp = m ^ xorArr[i];
 
            // If above XOR exists in map, then there
            // is another previous prefix with same
            // XOR, i.e., there is a subarray ending
            // at i with XOR equal to m.
            ans = ans
                  + (mp.ContainsKey(tmp) == false
                         ? 0
                         : ((long)mp[tmp]));
 
            // If this subarray has XOR equal to m itself.
            if (xorArr[i] == m)
                ans++;
 
            // Add the XOR of this subarray to the map
            if (mp.ContainsKey(xorArr[i]))
                mp[xorArr[i]] = mp[xorArr[i]] + 1;
            else
                mp.Add(xorArr[i], 1);
        }
 
        // Return total count of subarrays having XOR of
        // elements as given value m
        return ans;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 4, 2, 2, 6, 4 };
        int n = arr.Length;
        int m = 6;
 
        Console.Write(
            "Number of subarrays having given XOR is "
            + subarrayXor(arr, n, m));
    }
}
 
// This code is contributed by Rajput-Ji


Javascript




<script>
// Javascript Program to count all subarrays having
// XOR of elements as given value m with
// O(n) time complexity.
     
    // Returns count of subarrays of arr with XOR
    // value equals to m
    function subarrayXor(aee,n,m)
    {
        let ans = 0; // Initialize answer to be returned
  
        // Create a prefix xor-sum array such that
        // xorArr[i] has value equal to XOR
        // of all elements in arr[0 ..... i]
        let xorArr = new Array(n);
  
        // Create map that stores number of prefix array
        // elements corresponding to a XOR value
        let mp = new Map();
  
        // Initialize first element of prefix array
        xorArr[0] = arr[0];
  
        // Computing the prefix array.
        for (let i = 1; i < n; i++)
            xorArr[i] = xorArr[i - 1] ^ arr[i];
  
        // Calculate the answer
        for (let i = 0; i < n; i++) {
            // Find XOR of current prefix with m.
            let tmp = m ^ xorArr[i];
  
            // If above XOR exists in map, then there
            // is another previous prefix with same
            // XOR, i.e., there is a subarray ending
            // at i with XOR equal to m.
            ans = ans
                  + (mp.has(tmp) == false
                         ? 0
                         : (mp.get(tmp)));
  
            // If this subarray has XOR equal to m itself.
            if (xorArr[i] == m)
                ans++;
  
            // Add the XOR of this subarray to the map
            if (mp.has(xorArr[i]))
                mp.set(xorArr[i], mp.get(xorArr[i]) + 1);
            else
                mp.set(xorArr[i], 1);
        }
  
        // Return total count of subarrays having XOR of
        // elements as given value m
        return ans;
    }
     
     // Driver code
    let arr=[4, 2, 2, 6, 4];
     
    let n = arr.length;
    let m = 6;
    document.write("Number of subarrays having given XOR is "
            + subarrayXor(arr, n, m));
     
// This code is contributed by unknown2108
</script>


Output

Number of subarrays having given XOR is 4

Time Complexity: O(n)

Auxiliary Space: O(n)

Alternate Approach: Using Python Dictionary to store Prefix XOR

Implementation:

C++




#include <bits/stdc++.h>
 
using namespace std;
 
 
//Function to return the XOR of all subarrays
int subarrayXor(int arr[], int n, int m)
{
      //declaring the hashtable
      //and initializing it with a count of 1
      //for 0
    unordered_map <int, int> HashTable;
    HashTable[0] = 1;
    int count = 0, curSum = 0;
    for (int i = 0; i < n; i++)
    {
        curSum ^= arr[i];
        if (HashTable[curSum ^ m] > 0)
            count += HashTable[curSum ^ m];
        HashTable[curSum]++;
    }
    return(count);
}
         
 
// Driver program to test above function
int main()
{
    int arr[] = { 5, 6, 7, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int m = 5;
     
    //Function call
    cout << "Number of subarrays having given XOR is " << subarrayXor(arr, n, m);
}
 
 
// This code is contributed by phasing17


Java




// Java program to implement the approach
import java.util.*;
 
class GFG
{
 
  // Function to return the XOR of all subarrays
  static int subarrayXor(int[] arr, int n, int m)
  {
 
    // declaring the hashtable
    // and initializing it with a count of 1
    // for 0
    HashMap<Integer, Integer> HashTable
      = new HashMap<>();
 
    HashTable.put(0, 1);
    int count = 0, curSum = 0;
    for (int i = 0; i < n; i++) {
      curSum ^= arr[i];
      if (HashTable.containsKey(curSum ^ m))
        count += HashTable.get(curSum ^ m);
      if (!HashTable.containsKey(curSum))
        HashTable.put(curSum, 0);
      HashTable.put(curSum,
                    HashTable.get(curSum) + 1);
    }
    return (count);
  }
 
  // Driver program to test above function
  public static void main(String[] args)
  {
    int[] arr = { 5, 6, 7, 8, 9 };
    int n = arr.length;
    int m = 5;
 
    // Function call
    System.out.println(
      "Number of subarrays having given XOR is "
      + subarrayXor(arr, n, m));
  }
}
 
// This code is contributed by phasing17


Python3




from collections import defaultdict
def subarrayXor(arr, n, m):
    HashTable=defaultdict(bool)
    HashTable[0]=1
    count=0
    curSum=0
    for i in arr:
        curSum^=i
        if HashTable[curSum^m]:
            count+=HashTable[curSum^m]
        HashTable[curSum]+=1
    return(count)
         
 
# Driver program to test above function
def main():
    arr = [ 5, 6, 7, 8, 9 ]
    n = len(arr)
    m = 5
 
    print("Number of subarrays having given XOR is "
        , subarrayXor(arr, n, m))
 
if __name__ == '__main__':
    main()
     
 # This code is contributed by mrmechanical26052000


Javascript




<script>
// JavaScript program to implement the approach
 
// Function that calculates the number of subarrays
// with xor equal to given number
function subarrayXor(arr, n, m)
{
    // declaring the HashTable object
    var HashTable = {};
    HashTable["0"] = 1;
    var count = 0;
    var curSum = 0;
     
    // iterating over the array
    for (var i of arr)
    {
        curSum ^= i;
         
        // updating count if the xor of subarray is m
        if (HashTable.hasOwnProperty((curSum ^ m).toString()))
        {
            count += HashTable[(curSum ^ m).toString()];
        }
        if (HashTable.hasOwnProperty((curSum).toString()))
            HashTable[curSum.toString()] += 1;
        else
            HashTable[curSum.toString()] = 1;
         
    }
     
    return count;
}
 
// Driver Code
var arr = [ 5, 6, 7, 8, 9 ] ;
var n = arr.length;
var m = 5;
 
// Function Call
console.log("Number of subarrays having given XOR is ", subarrayXor(arr, n, m));
 
// This code is contributed by phasing17
</script>


Output

Number of subarrays having given XOR is 2

Time Complexity: O(n)

Auxiliary Space: O(n)

This article is contributed by Anmol Ratnam. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


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