# Make all the elements of array even with given operations

Given an array arr[] of positive integers, find the minimum number of operations required to make all the array elements even where:

1. If there is an odd number then, increment the element and the next adjacent element by 1.
2. Each increment costs one operations.

Note: If there is any number in arr[] which is odd after all operations then, print -1.

Examples

Input: arr[] = {2, 3, 4, 5, 6}
Output: 4
Explanation:
Now the array becomes {2, 4, 5, 5, 6}.
Now the array becomes {2, 4, 6, 6, 6}.
The resultant array has all even numbers.
The total number of operations for 4 increments is 4.

Input: arr[] = {5, 6}
Output: -1
Explanation:
Adding 1 to 5(0th index) then we have to increment 1 to its adjacent element 6(1st index).
Now the array becomes {6, 7}.
And we have 1 odd number left after all possible increments. Therefore we can’t make all array elements even.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
This problem can be solved using Greedy Approach. Following are the steps:

1. Traverse the given array arr[].
2. If an odd element occurs then increment that element by 1 to make it even and the next adjacent element by 1.
3. Repeat the above step for all the odd element for the given array arr[].
4. If all the element in arr[] are even then print the number of operation.
5. Else print -1.

Below is the implementation of the above approach:

## C++

 `// C++ program to make all array ` `// element even ` `#include "bits/stdc++.h" ` `using` `namespace` `std; ` ` `  `// Function to count the total ` `// number of operations needed to make ` `// all array element even ` `int` `countOperations(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``// Traverse the given array ` `    ``for` `(``int` `i = 0; i < n - 1; i++) { ` ` `  `        ``// If an odd element occurs ` `        ``// then increment that element ` `        ``// and next adjacent element ` `        ``// by 1 ` `        ``if` `(arr[i] & 1) { ` `            ``arr[i]++; ` `            ``arr[i + 1]++; ` `            ``count += 2; ` `        ``} ` `    ``} ` ` `  `    ``// Traverse the array if any odd ` `    ``// element occurs then return -1 ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(arr[i] & 1) ` `            ``return` `-1; ` `    ``} ` ` `  `    ``// Returns the count of operations ` `    ``return` `count; ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 3, 4, 5, 6 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` `    ``cout << countOperations(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to make all array ` `// element even ` `class` `GFG ` `{ ` ` `  `// Function to count the total ` `// number of operations needed to make ` `// all array element even ` `static` `int` `countOperations(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `count = ``0``; ` ` `  `    ``// Traverse the given array ` `    ``for` `(``int` `i = ``0``; i < n - ``1``; i++) ` `    ``{ ` ` `  `        ``// If an odd element occurs ` `        ``// then increment that element ` `        ``// and next adjacent element ` `        ``// by 1 ` `        ``if` `(arr[i] % ``2` `== ``1``)  ` `        ``{ ` `            ``arr[i]++; ` `            ``arr[i + ``1``]++; ` `            ``count += ``2``; ` `        ``} ` `    ``} ` ` `  `    ``// Traverse the array if any odd ` `    ``// element occurs then return -1 ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``if` `(arr[i] % ``2` `== ``1``) ` `            ``return` `-``1``; ` `    ``} ` ` `  `    ``// Returns the count of operations ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``2``, ``3``, ``4``, ``5``, ``6` `}; ` `    ``int` `n = arr.length; ` `    ``System.out.print(countOperations(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 program to make all array  ` `# element even  ` ` `  `# Function to count the total  ` `# number of operations needed to make  ` `# all array element even  ` `def` `countOperations(arr, n) : ` ` `  `    ``count ``=` `0``;  ` ` `  `    ``# Traverse the given array  ` `    ``for` `i ``in` `range``(n ``-` `1``) : ` ` `  `        ``# If an odd element occurs  ` `        ``# then increment that element  ` `        ``# and next adjacent element  ` `        ``# by 1  ` `        ``if` `(arr[i] & ``1``) : ` `            ``arr[i] ``+``=` `1``;  ` `            ``arr[i ``+` `1``] ``+``=` `1``;  ` `            ``count ``+``=` `2``;  ` ` `  `    ``# Traverse the array if any odd  ` `    ``# element occurs then return -1  ` `    ``for` `i ``in` `range``(n) : ` `        ``if` `(arr[i] & ``1``) : ` `            ``return` `-``1``;  ` ` `  `    ``# Returns the count of operations  ` `    ``return` `count;  ` ` `  `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``arr ``=` `[ ``2``, ``3``, ``4``, ``5``, ``6` `];  ` `    ``n ``=` `len``(arr);  ` `    ``print``(countOperations(arr, n));  ` `     `  `    ``# This code is contributed by AnkitRai01 `

## C#

 `// C# program to make all array ` `// element even ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to count the total ` `// number of operations needed to make ` `// all array element even ` `static` `int` `countOperations(``int` `[]arr, ``int` `n) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``// Traverse the given array ` `    ``for` `(``int` `i = 0; i < n - 1; i++) ` `    ``{ ` ` `  `        ``// If an odd element occurs ` `        ``// then increment that element ` `        ``// and next adjacent element ` `        ``// by 1 ` `        ``if` `(arr[i] % 2 == 1)  ` `        ``{ ` `            ``arr[i]++; ` `            ``arr[i + 1]++; ` `            ``count += 2; ` `        ``} ` `    ``} ` ` `  `    ``// Traverse the array if any odd ` `    ``// element occurs then return -1 ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``if` `(arr[i] % 2 == 1) ` `            ``return` `-1; ` `    ``} ` ` `  `    ``// Returns the count of operations ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = { 2, 3, 4, 5, 6 }; ` `    ``int` `n = arr.Length; ` `    ``Console.Write(countOperations(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```4
```

Time Complexity: O(N) where N is the number of elements in array.

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Improved By : 29AjayKumar, AnkitRai01

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