# Maximum even numbers present in any subarray of size K

• Difficulty Level : Medium
• Last Updated : 29 Apr, 2021

Given an array arr[] of size N and an integer K, the task is to find the maximum number of even numbers present in any subarray of size K.

Examples:

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Input: arr[] = {2, 3, 5, 4, 7, 6}, K = 3
Output:
Explanation:
Subarrays of size K(=3) with maximum count of even numbers are { arr, arr, arr }
Therefore, the required output is 2

Input: arr[] = {4, 3, 2, 6}, K = 2
Output: 2

Naive Approach: The simplest approach to solve this problem is to generate all possible subarrays of size K and count the even numbers in the subarray. Finally, print the maximum count obtained.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to find the maximum count of``// even numbers from all the subarrays of``// size K``int` `maxEvenIntegers(``int` `arr[], ``int` `N, ``int` `M)``{` `    ``// Stores the maximum count of even numbers``    ``// from all the subarrays of size K``    ``int` `ans = 0;` `    ``// Generate all subarrays of size K``    ``for` `(``int` `i = 0; i <= N - M; i++) {` `        ``// Store count of even numbers``        ``// in current subarray of size K``        ``int` `cnt = 0;` `        ``// Traverse the current subarray``        ``for` `(``int` `j = 0; j < M; j++) {` `            ``// If current element``            ``// is an even number``            ``if` `(arr[i + j] % 2 == 0)``                ``cnt++;``        ``}` `        ``// Update the answer``        ``ans = max(ans, cnt);``    ``}` `    ``// Return answer``    ``return` `ans;``}` `// Driver Code``int` `main()``{` `    ``int` `arr[] = { 2, 3, 5, 4, 7, 6 };``    ``int` `K = 3;` `    ``// Size of the input array``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << maxEvenIntegers(arr, N, K) << endl;` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;``class` `GFG``{` `// Function to find the maximum count of``// even numbers from all the subarrays of``// size K``static` `int` `maxEvenIntegers(``int` `arr[], ``int` `N, ``int` `M)``{` `    ``// Stores the maximum count of even numbers``    ``// from all the subarrays of size K``    ``int` `ans = ``0``;` `    ``// Generate all subarrays of size K``    ``for` `(``int` `i = ``0``; i <= N - M; i++)``    ``{` `        ``// Store count of even numbers``        ``// in current subarray of size K``        ``int` `cnt = ``0``;` `        ``// Traverse the current subarray``        ``for` `(``int` `j = ``0``; j < M; j++)``        ``{` `            ``// If current element``            ``// is an even number``            ``if` `(arr[i + j] % ``2` `== ``0``)``                ``cnt++;``        ``}` `        ``// Update the answer``        ``ans = Math.max(ans, cnt);``    ``}` `    ``// Return answer``    ``return` `ans;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``2``, ``3``, ``5``, ``4``, ``7``, ``6` `};``    ``int` `K = ``3``;` `    ``// Size of the input array``    ``int` `N = arr.length;``    ``System.out.print(maxEvenIntegers(arr, N, K) +``"\n"``);` `}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to find the maximum count of``# even numbers from all the subarrays of``# size K``def` `maxEvenIntegers(arr, N, K):``  ` `    ``# Stores the maximum count of even numbers``    ``# from all the subarrays of size K``    ``ans ``=` `0``    ``# Generate all subarrays of size K``    ``for` `i ``in` `range``(N``-``K``+``1``):``        ``# Store count of even numbers``        ``# in current subarray of size K``        ``cnt ``=` `0` `        ``# Traverse the current subarray``        ``for` `j ``in` `range``(``0``, K):``            ``if` `arr[i``+``j] ``%` `2` `=``=` `0``:``                ``cnt ``+``=` `1``        ``# Update the answer``        ``ans ``=` `max``(cnt, ans)``    ``# Return answer``    ``return` `ans`  `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``2``, ``3``, ``5``, ``4``, ``7``, ``6``]``    ``K ``=` `3``    ``# Size of the input array``    ``N ``=` `len``(arr)``    ``print``(maxEvenIntegers(arr, N, K))` `# This code is contributed by MuskanKalra1`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG``{``  ` `    ``// Function to find the maximum count of``    ``// even numbers from all the subarrays of``    ``// size K``    ``static` `int` `maxEvenIntegers(``int` `[]arr, ``int` `N, ``int` `M)``    ``{``    ` `        ``// Stores the maximum count of even numbers``        ``// from all the subarrays of size K``        ``int` `ans = 0;``    ` `        ``// Generate all subarrays of size K``        ``for` `(``int` `i = 0; i <= N - M; i++)``        ``{``    ` `            ``// Store count of even numbers``            ``// in current subarray of size K``            ``int` `cnt = 0;``    ` `            ``// Traverse the current subarray``            ``for` `(``int` `j = 0; j < M; j++)``            ``{``    ` `                ``// If current element``                ``// is an even number``                ``if` `(arr[i + j] % 2 == 0)``                    ``cnt++;``            ``}``    ` `            ``// Update the answer``            ``ans = Math.Max(ans, cnt);``        ``}``    ` `        ``// Return answer``        ``return` `ans;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``int` `[]arr = { 2, 3, 5, 4, 7, 6 };``        ``int` `K = 3;``    ` `        ``// Size of the input array``        ``int` `N = arr.Length;``        ``Console.WriteLine(maxEvenIntegers(arr, N, K));``    ``}``}` `// This code is contributed by AnkThon`

## Javascript

 ``
Output:
`2`

Time Complexity: O(N * K)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using the Sliding window technique. Follow the steps below to solve the problems:

• Initialize a variable, say cntMaxEven, to store the maximum count of even numbers in a subarray of size K.
• Calculate the count of even numbers in the subarray { arr, … arr[K – 1] } and store it into cntMaxEven.
• Traverse the remaining subarrays of size K by iterating over the range [K, N – 1]. For every ith iteration remove the first element of the subarray and insert the current ith element of the array into the current subarray.
• Count the even numbers in the current subarray and update cntMaxEven to the maximum count of even numbers in the current subarray and cntMaxEven.
• Finally, print the value of cntMaxEven.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to find the maximum count of``// even numbers from all the subarrays of``// size K``int` `maxEvenIntegers(``int` `arr[], ``int` `N, ``int` `M)``{` `    ``// Stores the count of even numbers``    ``// in a subarray of size K``    ``int` `curr = 0;` `    ``// Calculate the count of even numbers``    ``// in the current subarray``    ``for` `(``int` `i = 0; i < M; i++) {` `        ``// If current element is``        ``// an even number``        ``if` `(arr[i] % 2 == 0)``            ``curr++;``    ``}` `    ``// Stores the maximum count of even numbers``    ``// from all the subarrays of size K``    ``int` `ans = curr;` `    ``// Traverse remaining subarrays of size K``    ``// using sliding window technique``    ``for` `(``int` `i = M; i < N; i++) {` `        ``// If the first element of``        ``// the subarray is even``        ``if` `(arr[i - M] % 2 == 0) {` `            ``// Update curr``            ``curr--;``        ``}` `        ``// If i-th element is even increment``        ``// the count``        ``if` `(arr[i] % 2 == 0)``            ``curr++;` `        ``// Update the answer``        ``ans = max(ans, curr);``    ``}` `    ``// Return answer``    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 3, 5, 4, 7, 6 };``    ``int` `M = 3;` `    ``// Size of the input array``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``// Function call``    ``cout << maxEvenIntegers(arr, N, M) << endl;` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG``{` `  ``// Function to find the maximum count of``  ``// even numbers from all the subarrays of``  ``// size K``  ``static` `int` `maxEvenIntegers(``int` `arr[], ``int` `N, ``int` `M)``  ``{` `    ``// Stores the count of even numbers``    ``// in a subarray of size K``    ``int` `curr = ``0``;` `    ``// Calculate the count of even numbers``    ``// in the current subarray``    ``for` `(``int` `i = ``0``; i < M; i++)``    ``{` `      ``// If current element is``      ``// an even number``      ``if` `(arr[i] % ``2` `== ``0``)``        ``curr++;``    ``}` `    ``// Stores the maximum count of even numbers``    ``// from all the subarrays of size K``    ``int` `ans = curr;` `    ``// Traverse remaining subarrays of size K``    ``// using sliding window technique``    ``for` `(``int` `i = M; i < N; i++)``    ``{` `      ``// If the first element of``      ``// the subarray is even``      ``if` `(arr[i - M] % ``2` `== ``0``)``      ``{` `        ``// Update curr``        ``curr--;``      ``}` `      ``// If i-th element is even increment``      ``// the count``      ``if` `(arr[i] % ``2` `== ``0``)``        ``curr++;` `      ``// Update the answer``      ``ans = Math.max(ans, curr);``    ``}` `    ``// Return answer``    ``return` `ans;``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int` `arr[] = { ``2``, ``3``, ``5``, ``4``, ``7``, ``6` `};``    ``int` `M = ``3``;` `    ``// Size of the input array``    ``int` `N = arr.length;` `    ``// Function call``    ``System.out.print(maxEvenIntegers(arr, N, M) +``"\n"``);` `  ``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to find the maximum count of``# even numbers from all the subarrays of``# size M``def` `maxEvenIntegers(arr, N, M):``  ` `    ``# Stores the count of even numbers``    ``# in a subarray of size M``    ``curr ``=` `0``    ` `    ``# Calculate the count of even numbers``    ``# in the current subarray``    ``for` `i ``in` `range``(``0``, M):``      ` `        ``# If current element is``        ``# an even number``        ``if``(arr[i] ``%` `2` `=``=` `0``):``            ``curr ``+``=` `1``            ` `    ``# Stores the maximum count of even numbers``    ``# from all the subarrays of size M``    ``ans ``=` `curr``    ` `    ``# Traverse remaining subarrays of size M``    ``# using sliding window technique``    ``for` `i ``in` `range``(M, N):``      ` `        ``# If the first element of``        ``# the subarray is even``        ``if``(arr[i ``-` `M] ``%` `2` `=``=` `0``):``          ` `            ``# update curr``            ``curr ``-``=` `1``            ` `        ``# If i-th element is even increment``        ``# the count``        ``if``(arr[i] ``%` `2` `=``=` `0``):``            ``curr ``+``=` `1``            ` `            ``# update the answer``            ``ans ``=` `max``(curr, ans)``            ` `    ``# Return answer``    ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``2``, ``3``, ``5``, ``4``, ``7``, ``6``]``    ``M ``=` `3``    ` `    ``# Size of the input array``    ``N ``=` `len``(arr)``    ` `    ``# Function call``    ``print``(maxEvenIntegers(arr, N, M))` `# This code is contributed by MuskanKalra1`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG``{` `  ``// Function to find the maximum count of``  ``// even numbers from all the subarrays of``  ``// size K``  ``static` `int` `maxEvenints(``int` `[]arr, ``int` `N, ``int` `M)``  ``{` `    ``// Stores the count of even numbers``    ``// in a subarray of size K``    ``int` `curr = 0;` `    ``// Calculate the count of even numbers``    ``// in the current subarray``    ``for` `(``int` `i = 0; i < M; i++)``    ``{` `      ``// If current element is``      ``// an even number``      ``if` `(arr[i] % 2 == 0)``        ``curr++;``    ``}` `    ``// Stores the maximum count of even numbers``    ``// from all the subarrays of size K``    ``int` `ans = curr;` `    ``// Traverse remaining subarrays of size K``    ``// using sliding window technique``    ``for` `(``int` `i = M; i < N; i++)``    ``{` `      ``// If the first element of``      ``// the subarray is even``      ``if` `(arr[i - M] % 2 == 0)``      ``{` `        ``// Update curr``        ``curr--;``      ``}` `      ``// If i-th element is even increment``      ``// the count``      ``if` `(arr[i] % 2 == 0)``        ``curr++;` `      ``// Update the answer``      ``ans = Math.Max(ans, curr);``    ``}` `    ``// Return answer``    ``return` `ans;``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(String[] args)``  ``{``    ``int` `[]arr = { 2, 3, 5, 4, 7, 6 };``    ``int` `M = 3;` `    ``// Size of the input array``    ``int` `N = arr.Length;` `    ``// Function call``    ``Console.Write(maxEvenints(arr, N, M) +``"\n"``);``  ``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`2`

Time Complexity: O(N)

Auxiliary Space: O(1)

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