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.

// 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; }

Output:

Number of subarrays having given XOR is 4

Time Complexity of above solution is O(n^{2}).

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++ 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; }

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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.