# Count of ways to make Array sum even by removing only one element

Given an array arr[] positive integers, the task is to find the number of ways to convert the array sum even if we are allowed to remove only one element.

Examples:

Input: arr[] = { 1, 3, 3, 2 }
Output: 3
Explanation:
1. Remove 1, then sum is 3 + 3 + 2 = 8.
2. Remove 3, then sum is 1 + 3 + 2 = 6.
3. Remove 3, then sum is 1 + 3 + 2 = 6.

Input: arr[] = { 4, 8, 3, 3, 6 }
Output: 3
Explanation:
1. Remove 4, then sum is 8 + 3 + 3 + 6 = 20.
2. Remove 8, then sum is 4 + 3 + 3 + 6 = 16.
3. Remove 6, then sum is 4 + 8 + 3 + 3 = 18.

Approach: The key observation to the above problem statement is:

1. If we have an odd number of odd elements then the sum is always odd then we have to remove one odd number from the array arr[] to make the sum even. Since we have to remove one element, therefore, the total number of ways of making the sum even is the count of odd elements in the array arr[].
2. If we have an even number of odd elements then the sum is always even. Since we have to remove one element to make the sum even, therefore, the total number of ways of making the sum even is the count of even elements in the array arr[]

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find a number of ways ` `// to make array element sum even by ` `// removing one element ` `int` `find_num_of_ways(``int` `arr[], ``int` `N) ` `{ ` `    ``int` `count_even = 0, count_odd = 0; ` ` `  `    ``// Finding the count of even ` `    ``// and odd elements ` `    ``for` `(``int` `i = 0; i < N; i++) { ` `        ``if` `(arr[i] % 2) { ` `            ``count_odd++; ` `        ``} ` `        ``else` `{ ` `            ``count_even++; ` `        ``} ` `    ``} ` ` `  `    ``// If count_odd is odd then ` `    ``// no. of ways is count_odd ` `    ``if` `(count_odd % 2) { ` `        ``return` `count_odd; ` `    ``} ` ` `  `    ``// Else no. of ways is count_even ` `    ``else` `{ ` `        ``return` `count_even; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``// Given array arr[] ` `    ``int` `arr[] = { 1, 3, 3, 2 }; ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``// Function Call ` `    ``cout << find_num_of_ways(arr, N); ` `    ``return` `0; ` `} `

Output:

```3
```

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

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