Find number of subarrays with XOR value a power of 2

Given an integer array, arr[] of size N. The XOR value of any subarray of arr[] is defined as the xor of all the integers in that subarray. The task is to find the number of sub-arrays with XOR value a power of 2. (1, 2, 4, 8, 16, ….)
Examples:

Input :  arr[] = {2, 6, 7, 5, 8}
Output : 6
Subarrays : {2, 6}, {2}, {6, 7},  {6, 7, 5}, {7, 5}, {8}

Input : arr[] = {2, 4, 8, 16} 
Output : 4 

Approach : 

1. Maintain a hashmap which stores all the prefix XOR values and their counts.
2. At any index i, we need to find the number of subarrays which end at i and have XOR value a power of 2. We need to find for all the power of 2, the number of subarrays possible. For example. : Suppose current XOR value till index i is X, we need to find the number of subarrays which result in 16 (say S), so we need the count of prefix XOR Y such that (X^Y) = S or Y = S^X. Y can be find using Hash Map.
3. Perform Step 2, for all the index, add the output.

Below is the implementation of the above approach: 

C++

// C++ Program to count number of subarrays
// with Bitwise-XOR as power of 2

#include <bits/stdc++.h>
#define ll long long int
#define MAX 10
using namespace std;

// Function to find number of subarrays
int findSubarray(int array[], int n)
{
    // Hash Map to store prefix XOR values
    unordered_map<int, int> mp;

    // When no element is selected
    mp.insert({ 0, 1 });

    int answer = 0;
    int preXor = 0;
    for (int i = 0; i < n; i++) {
        int value = 1;
        preXor ^= array[i];

        // Check for all the powers of 2,
        // till a MAX value
        for (int j = 1; j <= MAX; j++) {
            int Y = value ^ preXor;
            if (mp.find(Y) != mp.end()) {
                answer += mp[Y];
            }
            value *= 2;
        }

        // Insert Current prefixxor in Hash Map
        if (mp.find(preXor) != mp.end()) {
            mp[preXor]++;
        }
        else {
            mp.insert({ preXor, 1 });
        }
    }

    return answer;
}

// Driver Code
int main()
{
    int array[] = { 2, 6, 7, 5, 8 };

    int n = sizeof(array) / sizeof(array[0]);

    cout << findSubarray(array, n) << endl;

    return 0;
}

Java

// Java Program to count number of subarrays
// with Bitwise-XOR as power of 2
import java.util.*;

class GFG
{

static int MAX = 10;

// Function to find number of subarrays
static int findSubarray(int array[], int n)
{
    // Hash Map to store prefix XOR values
    HashMap<Integer,
            Integer> mp = new HashMap<Integer,
                                      Integer>();

    // When no element is selected
    mp.put(0, 1);

    int answer = 0;
    int preXor = 0;
    for (int i = 0; i < n; i++)
    {
        int value = 1;
        preXor ^= array[i];

        // Check for all the powers of 2,
        // till a MAX value
        for (int j = 1; j <= MAX; j++)
        {
            int Y = value ^ preXor;
            if (mp.containsKey(Y))
            {
                answer += mp.get(Y);
            }
            value *= 2;
        }

        // Insert Current prefixxor in Hash Map
        if (mp.containsKey(preXor))
        {
            mp.put(preXor,mp.get(preXor) + 1);
        }
        else
        {
            mp.put(preXor, 1);
        }
    }
    return answer;
}

// Driver Code
public static void main (String[] args)
{
    int array[] = { 2, 6, 7, 5, 8 };

    int n = array.length;

    System.out.println(findSubarray(array, n));
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 Program to count number of subarrays
# with Bitwise-XOR as power of 2
MAX = 10

# Function to find number of subarrays
def findSubarray(array, n):

    # Hash Map to store prefix XOR values
    mp = dict()

    # When no element is selected
    mp[0] = 1

    answer = 0
    preXor = 0
    for i in range(n):
        value = 1
        preXor ^= array[i]

        # Check for all the powers of 2,
        # till a MAX value
        for j in range(1, MAX + 1):
            Y = value ^ preXor
            if ( Y in mp.keys()):
                answer += mp[Y]

            value *= 2

        # Insert Current prefixxor in Hash Map
        if (preXor in mp.keys()):
            mp[preXor] += 1

        else:
            mp[preXor] = 1

    return answer

# Driver Code
array = [2, 6, 7, 5, 8]

n = len(array)

print(findSubarray(array, n))

# This code is contributed by Mohit Kumar

C#

// C# Program to count number of subarrays
// with Bitwise-XOR as power of 2
using System;
using System.Collections.Generic;

class GFG
{
static int MAX = 10;

// Function to find number of subarrays
static int findSubarray(int []array, int n)
{
    // Hash Map to store prefix XOR values
    Dictionary<int,
               int> mp = new Dictionary<int,
                                        int>();

    // When no element is selected
    mp.Add(0, 1);

    int answer = 0;
    int preXor = 0;
    for (int i = 0; i < n; i++)
    {
        int value = 1;
        preXor ^= array[i];

        // Check for all the powers of 2,
        // till a MAX value
        for (int j = 1; j <= MAX; j++)
        {
            int Y = value ^ preXor;
            if (mp.ContainsKey(Y))
            {
                answer += mp[Y];
            }
            value *= 2;
        }

        // Insert Current prefixxor in Hash Map
        if (mp.ContainsKey(preXor))
        {
            mp[preXor] = mp[preXor] + 1;
        }
        else
        {
            mp.Add(preXor, 1);
        }
    }
    return answer;
}

// Driver Code
public static void Main (String[] args)
{
    int []array = { 2, 6, 7, 5, 8 };

    int n = array.Length;

    Console.WriteLine(findSubarray(array, n));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>


// Javascript Program to count number of subarrays
// with Bitwise-XOR as power of 2

var MAX = 10;

// Function to find number of subarrays
function findSubarray(array, n)
{
    // Hash Map to store prefix XOR values
    var mp = new Map();

    // When no element is selected
    mp.set(0,1);

    var answer = 0;
    var preXor = 0;
    for (var i = 0; i < n; i++) {
        var value = 1;
        preXor ^= array[i];

        // Check for all the powers of 2,
        // till a MAX value
        for (var j = 1; j <= MAX; j++) {
            var Y = value ^ preXor;
            if (mp.has(Y)) {
                answer += mp.get(Y);
            }
            value *= 2;
        }

        // Insert Current prefixxor in Hash Map
        if (mp.has(preXor)) {
            mp.set(preXor, mp.get(preXor)+1);
        }
        else {
            mp.set(preXor,1);
        }
    }

    return answer;
}

// Driver Code
var array = [2, 6, 7, 5, 8 ];
var n = array.length;
document.write( findSubarray(array, n));

</script>

Output:

6

Time Complexity: O(n * MAX)

Auxiliary Space: O(n)