# Number of subarrays such that XOR of one half is equal to the other

• Difficulty Level : Hard
• Last Updated : 15 Nov, 2022

Given an array of N numbers, the task is to find the number of sub-arrays (size of the sub-array should be an even number) of the given array such that after dividing the sub-array in two equal halves, bitwise XOR of one half of the sub-array will be equal to bitwise XOR of the other half.

Examples:

```Input : N = 6, arr[] = {3, 2, 2, 3, 7, 6}
Output : 3
Valid sub-arrays are {3, 2, 2, 3}, {2, 2},
and {2, 3, 7, 6}

Input : N = 5, arr[] = {1, 2, 3, 4, 5}
Output : 1

Input : N = 3, arr[] = {42, 4, 2}
Output : 0```

Approach: If an array is divided into two equal halves and XOR of one half is equal to the other, it means that XOR of whole of the array should be 0, because A^A = 0. Now, to solve the above problem find the prefix XOR’s of all the elements of the given array starting from left.

Suppose, a sub-array starts from l and ends at r, then (r-l+1) should be even. Also, to find XOR of a given range (l, r) subtract prefix XOR at (l – 1) with prefix XOR at r.

Since, (r – l + 1) is even, hence, if r is even then l should be odd and vice versa. Now, divide your prefixes into two groups, one should be the group of prefixes of odd indexes and the other should be of even indexes. Now, start traversing the prefix array from left to right and see how many time this particular prefix A has already occurred in its respective group, i.e. prefixes at even indexes should be checked in even prefix group and prefixes at odd indexes in odd prefix group (because if r is even then (l-1) is also even, similar logic is applied if r is odd).

Below is the implementation of the above approach:

## C++

 `// C++ program to find number of subarrays such that``// XOR of one half is equal to the other` `#include ``using` `namespace` `std;` `// Function to find number of subarrays such that``// XOR of one half is equal to the other``int` `findSubarrCnt(``int` `arr[], ``int` `n)``{``    ``// Variables to store answer and current XOR's``    ``int` `ans = 0, XOR = 0;` `    ``// Array to store prefix XOR's``    ``int` `prefix[n];` `    ``for` `(``int` `i = 0; i < n; ++i) {` `        ``// Calculate XOR until this index``        ``XOR = XOR ^ arr[i];` `        ``// Store the XOR in prefix array``        ``prefix[i] = XOR;``    ``}` `    ``// Create groups for odd indexes and even indexes``    ``unordered_map<``int``, ``int``> oddGroup, evenGroup;` `    ``// Initialize occurrence of 0 in oddGroup as 1``    ``// because it will be used in case our``    ``// subarray has l = 0``    ``oddGroup[0] = 1;` `    ``for` `(``int` `i = 0; i < n; ++i) {` `        ``if` `(i & 1) {` `            ``// Check the frequency of current prefix``            ``// XOR in oddGroup and add it to the``            ``// answer``            ``ans += oddGroup[prefix[i]];` `            ``// Update the frequency``            ``++oddGroup[prefix[i]];``        ``}``        ``else` `{` `            ``// Check the frequency of current prefix``            ``// XOR in evenGroup and add it to the``            ``// answer``            ``ans += evenGroup[prefix[i]];` `            ``// Update the frequency``            ``++evenGroup[prefix[i]];``        ``}``    ``}` `    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``int` `N = 6;` `    ``int` `arr[] = { 3, 2, 2, 3, 7, 6 };` `    ``cout << findSubarrCnt(arr, N);` `    ``return` `0;``}`

## Java

 `// JAVA program to find number of subarrays such that``// XOR of one half is equal to the other``import` `java.util.*;` `class` `GFG``{``        ` `    ``// Function to find number of subarrays such that``    ``// XOR of one half is equal to the other``    ``static` `int` `findSubarrCnt(``int` `arr[], ``int` `n)``    ``{``        ``// Variables to store answer and current XOR's``        ``int` `ans = ``0``, XOR = ``0``;``    ` `        ``// Array to store prefix XOR's``        ``int` `prefix[] = ``new` `int``[n];``    ` `        ``for` `(``int` `i = ``0``; i < n; ++i)``        ``{``    ` `            ``// Calculate XOR until this index``            ``XOR = XOR ^ arr[i];``    ` `            ``// Store the XOR in prefix array``            ``prefix[i] = XOR;``        ``}``    ` `        ``// Create groups for odd indexes and even indexes``        ``HashMap evenGroup = ``new` `HashMap<>();``        ``HashMap oddGroup = ``new` `HashMap<>();``        `  `        ``// Initialize occurrence of 0 in oddGroup as 1``        ``// because it will be used in case our``        ``// subarray has l = 0``        ``oddGroup.put(``0``, ``1``);``    ` `        ``for` `(``int` `i = ``0``; i < n; ++i)``        ``{``    ` `            ``if` `(i % ``2``== ``1``)``            ``{``    ` `                ``// Check the frequency of current prefix``                ``// XOR in oddGroup and add it to the``                ``// answer``                ``if``(oddGroup.containsKey(prefix[i]))``                ``{``                    ``ans += oddGroup.get(prefix[i]);``    ` `                    ``// Update the frequency``                    ``oddGroup.put(prefix[i],oddGroup.get(prefix[i] + ``1``));``                ``}``                ``else``                ``{``                    ``oddGroup.put(prefix[i], ``1``);``                ``}``                ` `            ``}``            ``else``            ``{``    ` `                ``// Check the frequency of current prefix``                ``// XOR in evenGroup and add it to the``                ``// answer``                ``if``(evenGroup.containsKey(prefix[i]))``                ``{``                    ``ans += evenGroup.get(prefix[i]);``    ` `                    ``// Update the frequency``                    ``evenGroup.put(prefix[i],evenGroup.get(prefix[i] + ``1``));``                ``}``                ``else``                ``{``                    ``evenGroup.put(prefix[i], ``1``);``                ``}``            ``}``        ``}``    ` `        ``return` `ans;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main (String[] args)``    ``{``    ` `        ``int` `arr[] = { ``3``, ``2``, ``2``, ``3``, ``7``, ``6` `};``        ``int` `N = arr.length;``    ` `        ``System.out.println(findSubarrCnt(arr, N));``    ``}``}` `// This code is contributed by ihritik`

## Python3

 `# Python3 program to find number of subarrays``# such that XOR of one half is equal to the other` `# Function to find number of subarrays``# such that XOR of one half is equal``# to the other``def` `findSubarrCnt(arr, n) :``    ` `    ``# Variables to store answer``    ``# and current XOR's``    ``ans ``=` `0``; XOR ``=` `0``;` `    ``# Array to store prefix XOR's``    ``prefix ``=` `[``0``] ``*` `n;` `    ``for` `i ``in` `range``(n) :` `        ``# Calculate XOR until this index``        ``XOR ``=` `XOR ^ arr[i];` `        ``# Store the XOR in prefix array``        ``prefix[i] ``=` `XOR;``    ` `    ``# Create groups for odd indexes and``    ``# even indexes``    ``oddGroup ``=` `dict``.fromkeys(prefix, ``0``)``    ``evenGroup ``=` `dict``.fromkeys(prefix, ``0``)` `    ``# Initialize occurrence of 0 in oddGroup``    ``# as 1 because it will be used in case ``    ``# our subarray has l = 0``    ``oddGroup[``0``] ``=` `1``;` `    ``for` `i ``in` `range``(n) :` `        ``if` `(i & ``1``) :` `            ``# Check the frequency of current``            ``# prefix XOR in oddGroup and add``            ``# it to the answer``            ``ans ``+``=` `oddGroup[prefix[i]];` `            ``# Update the frequency``            ``oddGroup[prefix[i]] ``+``=` `1``;``        ` `        ``else` `:` `            ``# Check the frequency of current``            ``# prefix XOR in evenGroup and add``            ``# it to the answer``            ``ans ``+``=` `evenGroup[prefix[i]];` `            ``# Update the frequency``            ``evenGroup[prefix[i]] ``+``=` `1``;` `    ``return` `ans;` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``N ``=` `6``;` `    ``arr ``=` `[ ``3``, ``2``, ``2``, ``3``, ``7``, ``6` `];` `    ``print``(findSubarrCnt(arr, N));` `# This code is contributed by Ryuga`

## C#

 `// C# program to find number of subarrays such that``// XOR of one half is equal to the other``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``        ` `    ``// Function to find number of subarrays such that``    ``// XOR of one half is equal to the other``    ``static` `int` `findSubarrCnt(``int` `[] arr, ``int` `n)``    ``{``        ``// Variables to store answer and current XOR's``        ``int` `ans = 0, XOR = 0;``    ` `        ``// Array to store prefix XOR's``        ``int` `[] prefix = ``new` `int``[n];``    ` `        ``for` `(``int` `i = 0; i < n; ++i)``        ``{``    ` `            ``// Calculate XOR until this index``            ``XOR = XOR ^ arr[i];``    ` `            ``// Store the XOR in prefix array``            ``prefix[i] = XOR;``        ``}``    ` `        ``// Create groups for odd indexes and even indexes``        ``Dictionary<``int``, ``int``> evenGroup = ``new` `Dictionary<``int``, ``int``>();``        ``Dictionary<``int``, ``int``> oddGroup = ``new` `Dictionary<``int``, ``int``>();` `        ``// Initialize occurrence of 0 in oddGroup as 1``        ``// because it will be used in case our``        ``// subarray has l = 0``        ``oddGroup[0] = 1;``    ` `        ``for` `(``int` `i = 0; i < n; ++i)``        ``{``    ` `            ``if` `(i % 2== 1)``            ``{``    ` `                ``// Check the frequency of current prefix``                ``// XOR in oddGroup and add it to the``                ``// answer``                ``if``(oddGroup.ContainsKey(prefix[i]))``                ``{``                    ``ans += oddGroup[prefix[i]];``    ` `                    ``// Update the frequency``                    ``oddGroup[prefix[i]]++;``                ``}``                ``else``                ``{``                    ``oddGroup[prefix[i]] = 1;``                ``}``                ` `            ``}``            ``else``            ``{``    ` `                ``// Check the frequency of current prefix``                ``// XOR in evenGroup and add it to the``                ``// answer``                ``if``(evenGroup.ContainsKey(prefix[i]))``                ``{``                    ``ans += evenGroup[prefix[i]];``    ` `                    ``// Update the frequency``                    ``evenGroup[prefix[i]]++;``                ``}``                ``else``                ``{``                    ``evenGroup[prefix[i]] = 1;``                ``}``            ``}``        ``}``    ` `        ``return` `ans;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main ()``    ``{``    ` `        ``int` `[] arr = { 3, 2, 2, 3, 7, 6 };``        ` `        ``int` `N = arr.Length;``    ` `        ``Console.WriteLine(findSubarrCnt(arr, N));``    ``}``}` `// This code is contributed by ihritik`

## Javascript

 ``

Output:

`3`

Time Complexity: O(N), where N is the size of the given input array.
Auxiliary Space: O(N), where N is the size of the given input array.

My Personal Notes arrow_drop_up