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: 1
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: 0
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.
- 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.
- Check if K is greater than the length of arr[] then return 0.
- 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.
- 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.
- 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.
- 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<bits/stdc++.h>using namespace std;// Function to return count of// required subarraysint 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 Codeint main(){ int arr[] = { 2, 4, 4, 2, 2, 4 }; int K = 4; int N = sizeof(arr) / sizeof(arr[0]); 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 frequenciesimport 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 subarraysdef 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 Codeif __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 frequenciesusing System;class GFG{// Function to return count of// required subarraysstatic 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 Codepublic 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
<script>// Javascript program to count subarrays// of size K with all elements// having even frequencies// Function to return count of// required subarraysfunction 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 var start = 0; var i = 0; // Stores the count of // required subarrays var count = 0; // Stores Xor of the // current subarray. var currXor = arr[i]; 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]; 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]; start++; } // Return count return count;}// Driver Code var arr = [2, 4, 4, 2, 2, 4]; var K = 4; var N = arr.length; document.write(countSubarray(arr, K, N));</script> |
Output:
3
Time Complexity: O(N)
Space Complexity: O(1)
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