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}

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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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 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

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C#

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// 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.

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PHP

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<?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.
?>

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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++

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// 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;
}

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Python3

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# 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

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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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