# Minimum removals to make array sum odd

Given an array Arr[] of N integers. The task is to find minimum number of elements needed to be removed from the array, so that sum of remaining elements is odd. Considering there is atleast one odd number.

Examples:

```Input:  arr[] = {1, 2, 3, 4}
Output: 1
Remove 1 to make array sum odd.

Input: arr[] =  {4, 2, 3, 4}
Output: 0
```

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

Approach: The idea to solve this problem is to first recall the below properties of ODDs and EVENs:

• odd + odd = even
• odd + even = odd
• even + even = even
• odd * even = even
• even * even = even
• odd * odd = odd

As the sum of any number of even numbers is even. So the count of even doesn’t matter.
And the sum of odd number of odd number is always odd. So to make array sum odd, count of odd numbers must be odd.
So, count the number of odd numbers and check if the count is odd then no removals required. Else the count is even then 1 odd number needs to be removed.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find minimum removals ` `int` `findCount(``int` `arr[], ``int` `n) ` `{ ` `    ``// Count odd numbers ` `    ``int` `countOdd = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``if` `(arr[i] % 2 == 1) ` `            ``countOdd++; ` ` `  `    ``// If the counter is odd return 0 ` `    ``// otherwise return 1 ` `    ``if` `(countOdd % 2 == 0) ` `        ``return` `1; ` `    ``else` `        ``return` `0; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 5, 1 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << findCount(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach  ` `class` `GfG  ` `{  ` ` `  `// Function to find minimum removals  ` `static` `int` `findCount(``int` `arr[], ``int` `n)  ` `{  ` `    ``// Count odd numbers  ` `    ``int` `countOdd = ``0``;  ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``if` `(arr[i] % ``2` `== ``1``)  ` `            ``countOdd++;  ` ` `  `    ``// If the counter is odd return 0  ` `    ``// otherwise return 1  ` `    ``if` `(countOdd % ``2` `== ``0``)  ` `        ``return` `1``;  ` `    ``else` `        ``return` `0``;  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``int` `arr[] = { ``1``, ``2``, ``3``, ``5``, ``1` `};  ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.println(findCount(arr, n));  ` `} ` `}  ` ` `  `// This code is contributed by ` `// Prerna Saini `

## Python3

 `# Python3 implementation of the  ` `# above approach  ` ` `  `# Function to find minimum removals  ` `def` `findCount(arr, n) : ` `     `  `    ``# Count odd numbers  ` `    ``countOdd ``=` `0``;  ` `    ``for` `i ``in` `range``(n) :  ` `        ``if` `(arr[i] ``%` `2` `=``=` `1``) : ` `            ``countOdd ``+``=` `1``;  ` ` `  `    ``# If the counter is odd return 0  ` `    ``# otherwise return 1  ` `    ``if` `(countOdd ``%` `2` `=``=` `0``) :  ` `        ``return` `1``;  ` `    ``else` `: ` `        ``return` `0``;  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` `    ``arr ``=` `[ ``1``, ``2``, ``3``, ``5``, ``1` `];  ` `    ``n ``=` `len``(arr) ; ` ` `  `    ``print``(findCount(arr, n));  ` ` `  `# This code is contributed by Ryuga `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GfG  ` `{  ` ` `  `// Function to find minimum removals  ` `static` `int` `findCount(``int` `[]arr, ``int` `n)  ` `{  ` `    ``// Count odd numbers  ` `    ``int` `countOdd = 0;  ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``if` `(arr[i] % 2 == 1)  ` `            ``countOdd++;  ` ` `  `    ``// If the counter is odd return 0  ` `    ``// otherwise return 1  ` `    ``if` `(countOdd % 2 == 0)  ` `        ``return` `1;  ` `    ``else` `        ``return` `0;  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `Main(String[] args)  ` `{  ` `    ``int` `[]arr = { 1, 2, 3, 5, 1 };  ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.WriteLine(findCount(arr, n));  ` `} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## PHP

 ` `

Output:

```1
```

My Personal Notes arrow_drop_up Discovering ways to develop a plane for soaring career goals

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.