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# Maximize the count of adjacent element pairs with even sum by rearranging the Array

Given an array, arr[] of N integers, the task is to find the maximum possible count of adjacent pairs with an even sum, rearranging the array arr[].

Examples:

Input: arr[] = {5, 5, 1}
Output: 2
Explanation:
The given array is already arranged to give the maximum count of adjacent pairs with an even sum.

1. {arr(= 5), arr(= 5}, the sum of the elements is 10, which is even.
2. {arr(= 5), arr(= 1}, the sum of the elements is 6, which is even.

Therefore, there are totals of 2 adjacent pairs with an even sum. And it is also the maximum possible count.

Input: arr[] = {9, 13, 15, 3, 16, 9, 13, 18}
Output: 6
Explanation:
One way to obtain the maximum count is to rearrange the array as {9, 9, 3, 13, 13, 15, 16, 18}.

1. {arr(= 9), arr(= 9}, the sum of the elements is 18, which is even.
2. {arr(= 9), arr(= 3}, the sum of the elements is 12, which is even.
3. {arr(= 3), arr(= 13}, the sum of the elements is 16, which is even.
4. {arr(= 13), arr(= 13}, the sum of the elements is 26, which is even.
5. {arr(= 13), arr(= 15}, the sum of the elements is 28, which is even.
6. {arr(= 15), arr(= 16}, the sum of the elements is 31, which is not even.
7. {arr(= 16), arr(= 18}, the sum of the elements is 34, which is even.

Therefore, there are a total of 6 adjacent pairs with an even sum. And it is also the maximum possible count.

Naive Approach: The simplest approach is to try every possible arrangement of the elements and then count the number of the adjacent pairs with an even sum.

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

Efficient Approach: The above approach can be optimized based on the following observations:

1. It is known that:
• Odd + Odd = Even
• Even + Even = Even
• Even + Odd = Odd
• Odd + Even = Odd
2. The total count of adjacent pairs is N-1.
3. Therefore, the maximum count can be obtained by putting all even numbers together and then all odd numbers or vice versa.
4. Rearranging in the above-mentioned way, there will be only one pair of adjacent elements with an odd sum which will be at the junction of even numbers and odd numbers.

Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find maximum count``// pair of adjacent elements with``// even sum``int` `maximumCount(``int` `arr[], ``int` `N)``{``    ``// Stores count of odd numbers``    ``int` `odd = 0;` `    ``// Stores count of even numbers``    ``int` `even = 0;` `    ``// Traverse the array arr[]``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// If arr[i]%2 is 1``        ``if` `(arr[i] % 2)``            ``odd++;``        ``// Else``        ``else``            ``even++;``    ``}` `    ``// If odd and even both``    ``// are greater than 0``    ``if` `(odd and even)``        ``return` `N - 2;``    ``// Otherwise``    ``else``        ``return` `N - 1;``}` `// Driver Code``int` `main()` `{``    ``int` `arr[] = { 9, 13, 15, 3, 16, 9, 13, 18 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << maximumCount(arr, N);` `    ``return` `0;``}`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;` `class` `GFG {``    ``// Function to find maximum count``    ``// pair of adjacent elements with``    ``// even sum``    ``static` `int` `maximumCount(``int` `arr[], ``int` `N)``    ``{``        ``// Stores count of odd numbers``        ``int` `odd = ``0``;` `        ``// Stores count of even numbers``        ``int` `even = ``0``;` `        ``// Traverse the array arr[]``        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``// If arr[i]%2 is 1``            ``if` `(arr[i] % ``2` `== ``1``)``                ``odd++;``            ``// Else``            ``else``                ``even++;``        ``}` `        ``// If odd and even both``        ``// are greater than 0``        ``if` `(odd > ``0` `&& even > ``0``)``            ``return` `N - ``2``;``        ``// Otherwise``        ``else``            ``return` `N - ``1``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``9``, ``13``, ``15``, ``3``, ``16``, ``9``, ``13``, ``18` `};``        ``int` `N = arr.length;` `        ``System.out.println(maximumCount(arr, N));``    ``}``}` `    ``// This code is contributed by Potta Lokesh`

## Python3

 `# Python 3 program for the above approach` `# Function to find maximum count``# pair of adjacent elements with``# even sum``def` `maximumCount(arr, N):``  ` `    ``# Stores count of odd numbers``    ``odd ``=` `0` `    ``# Stores count of even numbers``    ``even ``=` `0` `    ``# Traverse the array arr[]``    ``for` `i ``in` `range``(N):``      ` `        ``# If arr[i]%2 is 1``        ``if` `(arr[i] ``%` `2``):``            ``odd ``+``=` `1``        ``# Else``        ``else``:``            ``even ``+``=` `1` `    ``# If odd and even both``    ``# are greater than 0``    ``if` `(odd ``and` `even):``        ``return` `N ``-` `2``    ``# Otherwise``    ``else``:``        ``return` `N ``-` `1` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``9``, ``13``, ``15``, ``3``, ``16``, ``9``, ``13``, ``18``]``    ``N ``=` `len``(arr)` `    ``print``(maximumCount(arr, N))``    ` `    ``# This code is contributed by bgangwar59.`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to find maximum count``// pair of adjacent elements with``// even sum``static` `int` `maximumCount(``int` `[]arr, ``int` `N)``{``    ``// Stores count of odd numbers``    ``int` `odd = 0;` `    ``// Stores count of even numbers``    ``int` `even = 0;` `    ``// Traverse the array arr[]``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// If arr[i]%2 is 1``        ``if` `(arr[i] % 2 !=0)``            ``odd++;``        ``// Else``        ``else``            ``even++;``    ``}` `    ``// If odd and even both``    ``// are greater than 0``    ``if` `(odd!=0 && even!=0)``        ``return` `N - 2;``    ``// Otherwise``    ``else``        ``return` `N - 1;``}` `// Driver Code``public` `static` `void` `Main()` `{``    ``int` `[]arr = { 9, 13, 15, 3, 16, 9, 13, 18 };``    ``int` `N = arr.Length;` `    ``Console.Write(maximumCount(arr, N));` `}``}` `// This code is contributed by ipg2016107.`

## Javascript

 ``

Output:

`6`

Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.

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