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# Make all array elements even by replacing any pair of array elements with their sum

• Last Updated : 28 Feb, 2022

Given an array arr[] consisting of N positive integers, the task is to make all array elements even by replacing any pair of array elements with their sum.

Examples:

Input: arr[] = {5, 6, 3, 7, 20}
Output: 3
Explanation:
Operation 1: Replace arr[0] and arr[2] by their sum ( = 5 + 3 = 8) modifies arr[] to {8, 6, 8, 7, 20}.
Operation 2: Replace arr[2] and arr[3] by their sum ( = 7 + 8 = 15) modifies arr[] to {8, 6, 15, 15, 20}.
Operation 3: Replace arr[2] and arr[3] by their sum ( = 15 + 15 = 30) modifies arr[] to {8, 6, 30, 30, 20}.

Input: arr[] = {2, 4, 16, 8, 7, 9, 3, 1}
Output: 2

Approach: The idea is to keep replacing two odd array elements by their sum until all array elements are even. 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 the minimum number` `// of replacements required to make` `// all array elements even` `void` `minMoves(``int` `arr[], ``int` `N)` `{` `    ``// Stores the count of odd elements` `    ``int` `odd_element_cnt = 0;`   `    ``// Traverse the array` `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``// Increase count of odd elements` `        ``if` `(arr[i] % 2 != 0) {` `            ``odd_element_cnt++;` `        ``}` `    ``}`   `    ``// Store number of replacements required` `    ``int` `moves = (odd_element_cnt) / 2;`   `    ``// Two extra moves will be required` `    ``// to make the last odd element even` `    ``if` `(odd_element_cnt % 2 != 0)` `        ``moves += 2;`   `    ``// Print the minimum replacements` `    ``cout << moves;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 5, 6, 3, 7, 20 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``// Function call` `    ``minMoves(arr, N);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.util.*;` `class` `GFG{`   `// Function to find the minimum number` `// of replacements required to make` `// all array elements even` `static` `void` `minMoves(``int` `arr[], ``int` `N)` `{` `  `  `    ``// Stores the count of odd elements` `    ``int` `odd_element_cnt = ``0``;`   `    ``// Traverse the array` `    ``for` `(``int` `i = ``0``; i < N; i++)` `    ``{`   `        ``// Increase count of odd elements` `        ``if` `(arr[i] % ``2` `!= ``0``) ` `        ``{` `            ``odd_element_cnt++;` `        ``}` `    ``}`   `    ``// Store number of replacements required` `    ``int` `moves = (odd_element_cnt) / ``2``;`   `    ``// Two extra moves will be required` `    ``// to make the last odd element even` `    ``if` `(odd_element_cnt % ``2` `!= ``0``)` `        ``moves += ``2``;`   `    ``// Print the minimum replacements` `    ``System.out.print(moves);` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `arr[] = { ``5``, ``6``, ``3``, ``7``, ``20` `};` `    ``int` `N = arr.length;`   `    ``// Function call` `    ``minMoves(arr, N);` `}` `}`   `// This code is contributed by shikhasingrajput`

## C#

 `// C# program for the above approach` `using` `System;` `public` `class` `GFG` `{`   `  ``// Function to find the minimum number` `  ``// of replacements required to make` `  ``// all array elements even` `  ``static` `void` `minMoves(``int` `[]arr, ``int` `N)` `  ``{`   `    ``// Stores the count of odd elements` `    ``int` `odd_element_cnt = 0;`   `    ``// Traverse the array` `    ``for` `(``int` `i = 0; i < N; i++)` `    ``{`   `      ``// Increase count of odd elements` `      ``if` `(arr[i] % 2 != 0) ` `      ``{` `        ``odd_element_cnt++;` `      ``}` `    ``}`   `    ``// Store number of replacements required` `    ``int` `moves = (odd_element_cnt) / 2;`   `    ``// Two extra moves will be required` `    ``// to make the last odd element even` `    ``if` `(odd_element_cnt % 2 != 0)` `      ``moves += 2;`   `    ``// Print the minimum replacements` `    ``Console.Write(moves);` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `Main(String[] args)` `  ``{` `    ``int` `[]arr = { 5, 6, 3, 7, 20 };` `    ``int` `N = arr.Length;`   `    ``// Function call` `    ``minMoves(arr, N);` `  ``}` `}`   `// This code is contributed by 29AjayKumar`

## Python3

 `# Python program for the above approach`   `# Function to find the minimum number` `# of replacements required to make` `# all array elements even` `def` `minMoves(arr, N):` `  `  `    ``# Stores the count of odd elements` `    ``odd_element_cnt ``=` `0``;`   `    ``# Traverse the array` `    ``for` `i ``in` `range``(N):`   `        ``# Increase count of odd elements` `        ``if` `(arr[i] ``%` `2` `!``=` `0``):` `            ``odd_element_cnt ``+``=` `1``;`   `    ``# Store number of replacements required` `    ``moves ``=` `(odd_element_cnt) ``/``/` `2``;`   `    ``# Two extra moves will be required` `    ``# to make the last odd element even` `    ``if` `(odd_element_cnt ``%` `2` `!``=` `0``):` `        ``moves ``+``=` `2``;`   `    ``# Print the minimum replacements` `    ``print``(moves);`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``5``, ``6``, ``3``, ``7``, ``20``];` `    ``N ``=` `len``(arr);`   `    ``# Function call` `    ``minMoves(arr, N);`   `    ``# This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`3`

Time complexity: O(N)
Auxiliary Space: O(1)

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