Count the number of subarrays having a given XOR

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 2 are {5}
              and {5, 6, 7, 8, 9}

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

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 po i of subarrays 
    for i in range(0,n): 
          
        xorSum = 0 # Store XOR of current subarray 
   
        # Pick ending po j of subarray for each i 
        for j  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. 
?> 

Output:

Number of subarrays having given XOR is 4

Time Complexity of above solution is O(n²).

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 sum of this count over all 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.

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 

Output:

Number of subarrays having given XOR is 4

Time Complexity: 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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

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