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# Count of subarrays of size K with elements having even frequencies

• Difficulty Level : Hard
• Last Updated : 12 May, 2021

Given an array arr[] and an integer K, the task is to count subarrays of size K in which every element appears an even number of times in the subarray.

Examples:

Input: arr[] = {1, 4, 2, 10, 2, 10, 0, 20}, K = 4
Output:
Explanation: Only subarray {2, 10, 2, 10} satisfies the required condition.

Input: arr[] = {1, 4, 2, 10, 2, 3, 1, 0, 20}, K = 3
Output:

Naive Approach:
The idea is to generate all subarrays of size K and check each of them whether all its elements are present even a number of times or not.

Time complexity: O(N*K)
Efficient Approach:

The idea is to use Window Sliding and the XOR concept here.

1. If the given K is odd, then return 0 as it is guaranteed that at least one number appears an odd number of times if K is odd.
2. Check if K is greater than the length of arr[] then return 0.
3. Initialize the following variables:
• count: Store the count of subarrays of size K with all elements.
• start: Remove left most element which is no longer part of k size subarray.
• currXor: Store Xor of the current subarray.
4. Calculate the Xor of the first K size subarray and check if currXor becomes 0, then increment the count and update currXor by eliminating Xor with arr[start] and increment start by 1.
5. Traverse arr[] from K to the length of arr[]:
• Update currXor by doing Xor with arr[i].
• Increment count if currXor becomes 0 otherwise ignore.
• Update currXor by eliminating Xor with arr[start].
• Increment start by 1.
6. Return count.

Below is the implementation of the above approach:

## C++

 `// C++ program to count subarrays``// of size K with all elements``// having even frequencies``#include``using` `namespace` `std;` `// Function to return count of``// required subarrays``int` `countSubarray(``int` `arr[], ``int` `K,``                             ``int` `N)``{``    ` `    ``// If K is odd``    ``if` `(K % 2 != 0)``        ` `        ``// Not possible to have``        ``// any such subarrays``        ``return` `0;` `    ``if` `(N < K)``        ``return` `0;` `    ``// Stores the starting index``    ``// of every subarrays``    ``int` `start = 0;` `    ``int` `i = 0;``    ` `    ``// Stores the count of``    ``// required subarrays``    ``int` `count = 0;``    ` `    ``// Stores Xor of the``    ``// current subarray.``    ``int` `currXor = arr[i++];` `    ``// Xor of first subarray``    ``// of size K``    ``while` `(i < K)``    ``{``        ``currXor ^= arr[i];``        ``i++;``    ``}` `    ``// If all elements appear``    ``// even number of times,``    ``// increase the count of``    ``// such subarrays``    ``if` `(currXor == 0)``        ``count++;` `    ``// Remove the starting element``    ``// from the current subarray``    ``currXor ^= arr[start++];` `    ``// Traverse the array``    ``// for the remaining``    ``// subarrays``    ``while` `(i < N)``    ``{``        ` `        ``// Update Xor by adding the``        ``// last element of the``        ``// current subarray``        ``currXor ^= arr[i];``        ` `        ``// Increment i``        ``i++;` `        ``// If currXor becomes 0,``        ``// then increment count``        ``if` `(currXor == 0)``            ``count++;` `        ``// Update currXor by removing``        ``// the starting element of the``        ``// current subarray``        ``currXor ^= arr[start++];``    ``}` `    ``// Return count``    ``return` `count;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 4, 4, 2, 2, 4 };``    ``int` `K = 4;``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ` `    ``cout << (countSubarray(arr, K, N));``}` `// This code is contributed by chitranayal`

## Java

 `// Java program to count subarrays``// of size K with all elements``// having even frequencies` `import` `java.util.*;` `class` `GFG {` `    ``// Function to return count of``    ``// required subarrays``    ``static` `int` `countSubarray(``int``[] arr,``                             ``int` `K, ``int` `N)``    ``{``        ``// If K is odd``        ``if` `(K % ``2` `!= ``0``)``            ``// Not possible to have``            ``// any such subarrays``            ``return` `0``;` `        ``if` `(N < K)``            ``return` `0``;` `        ``// Stores the starting index``        ``// of every subarrays``        ``int` `start = ``0``;` `        ``int` `i = ``0``;``        ``// Stores the count of``        ``// required subarrays``        ``int` `count = ``0``;``        ``// Stores Xor of the``        ``// current subarray.``        ``int` `currXor = arr[i++];` `        ``// Xor of first subarray``        ``// of size K``        ``while` `(i < K) {``            ``currXor ^= arr[i];``            ``i++;``        ``}` `        ``// If all elements appear``        ``// even number of times,``        ``// increase the count of``        ``// such subarrays``        ``if` `(currXor == ``0``)``            ``count++;` `        ``// Remove the starting element``        ``// from the current subarray``        ``currXor ^= arr[start++];` `        ``// Traverse the array``        ``// for the remaining``        ``// subarrays``        ``while` `(i < N) {``            ``// Update Xor by adding the``            ``// last element of the``            ``// current subarray``            ``currXor ^= arr[i];``            ``// Increment i``            ``i++;` `            ``// If currXor becomes 0,``            ``// then increment count``            ``if` `(currXor == ``0``)``                ``count++;` `            ``// Update currXor by removing``            ``// the starting element of the``            ``// current subarray``            ``currXor ^= arr[start++];``        ``}` `        ``// Return count``        ``return` `count;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = { ``2``, ``4``, ``4``, ``2``, ``2``, ``4` `};``        ``int` `K = ``4``;``        ``int` `N = arr.length;``        ``System.out.println(``            ``countSubarray(arr, K, N));``    ``}``}`

## Python3

 `# Python3 program to count subarrays``# of size K with all elements``# having even frequencies` `# Function to return count of``# required subarrays``def` `countSubarray(arr, K, N):``    ` `    ``# If K is odd``    ``if` `(K ``%` `2` `!``=` `0``):``        ` `        ``# Not possible to have``        ``# any such subarrays``        ``return` `0` `    ``if` `(N < K):``        ``return` `0` `    ``# Stores the starting index``    ``# of every subarrays``    ``start ``=` `0``    ``i ``=` `0``    ` `    ``# Stores the count of``    ``# required subarrays``    ``count ``=` `0``    ` `    ``# Stores Xor of the``    ``# current subarray.``    ``currXor ``=` `arr[i]``    ``i ``+``=` `1` `    ``# Xor of first subarray``    ``# of size K``    ``while` `(i < K):``        ``currXor ^``=` `arr[i]``        ``i ``+``=` `1` `    ``# If all elements appear``    ``# even number of times,``    ``# increase the count of``    ``# such subarrays``    ``if` `(currXor ``=``=` `0``):``        ``count ``+``=` `1` `    ``# Remove the starting element``    ``# from the current subarray``    ``currXor ^``=` `arr[start]``    ``start ``+``=` `1` `    ``# Traverse the array``    ``# for the remaining``    ``# subarrays``    ``while` `(i < N):``        ` `        ``# Update Xor by adding the``        ``# last element of the``        ``# current subarray``        ``currXor ^``=` `arr[i]``        ` `        ``# Increment i``        ``i ``+``=` `1` `        ``# If currXor becomes 0,``        ``# then increment count``        ``if` `(currXor ``=``=` `0``):``            ``count ``+``=` `1` `        ``# Update currXor by removing``        ``# the starting element of the``        ``# current subarray``        ``currXor ^``=` `arr[start]``        ``start ``+``=` `1` `    ``# Return count``    ``return` `count``    ` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``arr ``=` `[ ``2``, ``4``, ``4``, ``2``, ``2``, ``4` `]``    ``K ``=` `4``    ``N ``=` `len``(arr)``    ` `    ``print``(countSubarray(arr, K, N))` `# This code is contributed by mohit kumar 29   `

## C#

 `// C# program to count subarrays``// of size K with all elements``// having even frequencies``using` `System;``class` `GFG{` `// Function to return count of``// required subarrays``static` `int` `countSubarray(``int``[] arr,``                         ``int` `K, ``int` `N)``{``    ``// If K is odd``    ``if` `(K % 2 != 0)``    ` `        ``// Not possible to have``        ``// any such subarrays``        ``return` `0;` `    ``if` `(N < K)``        ``return` `0;` `    ``// Stores the starting index``    ``// of every subarrays``    ``int` `start = 0;` `    ``int` `i = 0;``    ` `    ``// Stores the count of``    ``// required subarrays``    ``int` `count = 0;``    ` `    ``// Stores Xor of the``    ``// current subarray.``    ``int` `currXor = arr[i++];` `    ``// Xor of first subarray``    ``// of size K``    ``while` `(i < K)``    ``{``        ``currXor ^= arr[i];``        ``i++;``    ``}` `    ``// If all elements appear``    ``// even number of times,``    ``// increase the count of``    ``// such subarrays``    ``if` `(currXor == 0)``        ``count++;` `    ``// Remove the starting element``    ``// from the current subarray``    ``currXor ^= arr[start++];` `    ``// Traverse the array``    ``// for the remaining``    ``// subarrays``    ``while` `(i < N)``    ``{``        ``// Update Xor by adding the``        ``// last element of the``        ``// current subarray``        ``currXor ^= arr[i];``        ` `        ``// Increment i``        ``i++;` `        ``// If currXor becomes 0,``        ``// then increment count``        ``if` `(currXor == 0)``            ``count++;` `        ``// Update currXor by removing``        ``// the starting element of the``        ``// current subarray``        ``currXor ^= arr[start++];``    ``}` `    ``// Return count``    ``return` `count;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int``[] arr = { 2, 4, 4, 2, 2, 4 };``    ``int` `K = 4;``    ``int` `N = arr.Length;``    ``Console.Write(countSubarray(arr, K, N));``}``}` `// This code is contributed by Akanksha_Rai`

## Javascript

 ``
Output:
`3`

Time Complexity: O(N)
Space Complexity: O(1)

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