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Minimum removals to make array sum odd
• Difficulty Level : Medium
• Last Updated : 29 Apr, 2021

Given an array Arr[] of N integers. The task is to find the minimum number of elements needed to be removed from the array so that the sum of the remaining elements is odd. Considering there is at least 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
Sum is already odd.```

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 the odd numbers of an odd number is always odd. So to make the array sum odd, the 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

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## Javascript

 ``
Output:
`1`

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