# Check if sum of exactly K elements of the Array can be odd or not

Given an array, arr[] and an integer K. Check whether it is possible to get an odd sum by choosing exactly K elements of the array.

Examples:

Input: arr[] = { 1, 2, 3 }, K = 2
Output: Possible
Explanation: {2, 3} -> 2 + 3 = 5

Input: arr[] = { 1, 2, 4, 2 }, K = 4
Output: Not Possible
Explanation: {2, 2, 4, 2} -> 2 + 2 + 4 + 2 = 10
No other possibilities as K is equal to the size of array

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

Approach:

On observing, it is found that there are three cases.

• First count the number of odd and even elements.
• Case 1: When all elements are even. Then, sum will always be even irrespective of the value of K as even + even = even.
• Case 2: When all elements are odd. Then, the sum will depend only on the value of K.

If K is odd, then the sum will be odd as every odd pair makes the sum even and in the end, one odd element makes the sum odd as even + odd = odd.
If K is even then every odd element get paired and become even, therefore the sum becomes even.

• Case 3: When K <= N, then sum depends only on the number of odd elements. If the number of odd elements is even, then the sum will be even as odd + odd = even, which implies every odd pair will become even. And if we add even elements to the sum, the sum will remain even as even + even = even.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function returns true if ` `// it is possible to have ` `// odd sum ` `bool` `isPossible(``int` `arr[], ` `                ``int` `N, ``int` `K) ` `{ ` `    ``int` `oddCount = 0, evenCount = 0; ` ` `  `    ``// counting number of odd ` `    ``// and even elements ` `    ``for` `(``int` `i = 0; i < N; i++) { ` `        ``if` `(arr[i] % 2 == 0) ` `            ``evenCount++; ` `        ``else` `            ``oddCount++; ` `    ``} ` `    ``if` `(evenCount == N ` `        ``|| (oddCount == N && K % 2 == 0) ` `        ``|| (K == N && oddCount % 2 == 0)) ` `        ``return` `false``; ` `    ``else` `        ``return` `true``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 4, 5, 8 }; ` `    ``int` `K = 5; ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``if` `(isPossible(arr, N, K)) ` `        ``cout << ``"Possible"``; ` `    ``else` `        ``cout << ``"Not Possible"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach  ` `class` `GFG{ ` ` `  `// Function returns true if  ` `// it is possible to have  ` `// odd sum  ` `static` `boolean` `isPossible(``int` `arr[],  ` `                          ``int` `N, ``int` `K)  ` `{  ` `    ``int` `oddCount = ``0``, evenCount = ``0``;  ` `     `  `    ``// Counting number of odd  ` `    ``// and even elements  ` `    ``for``(``int` `i = ``0``; i < N; i++) ` `    ``{  ` `       ``if` `(arr[i] % ``2` `== ``0``) ` `       ``{  ` `           ``evenCount++; ` `       ``}  ` `       ``else` `       ``{ ` `           ``oddCount++; ` `       ``} ` `    ``}  ` `    ``if` `(evenCount == N ||  ` `       ``(oddCount == N && K % ``2` `== ``0``) || ` `       ``(K == N && oddCount % ``2` `== ``0``)) ` `    ``{  ` `        ``return` `false``; ` `    ``}  ` `    ``else` `    ``{ ` `        ``return` `true``; ` `    ``}  ` `}  ` `     `  `// Driver code  ` `public` `static` `void` `main (String[] args)  ` `{  ` `    ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ``8` `};  ` `    ``int` `K = ``5``;  ` `    ``int` `N = arr.length;  ` `     `  `    ``if` `(isPossible(arr, N, K)) ` `    ``{ ` `        ``System.out.println(``"Possible"``); ` `    ``}  ` `    ``else` `    ``{ ` `        ``System.out.println(``"Not Possible"``); ` `    ``}  ` `}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the above approach ` ` `  `# Function returns true if it  ` `# is possible to have odd sum ` `def` `isPossible(arr, N, K): ` `     `  `    ``oddCount ``=` `0` `    ``evenCount ``=` `0` ` `  `    ``# Counting number of odd ` `    ``# and even elements ` `    ``for` `i ``in` `range``(N): ` `        ``if` `(arr[i] ``%` `2` `=``=` `0``): ` `            ``evenCount ``+``=` `1` `        ``else``: ` `            ``oddCount ``+``=` `1` `             `  `    ``if` `(evenCount ``=``=` `N ``or`  `       ``(oddCount ``=``=` `N ``and` `K ``%` `2` `=``=` `0``) ``or`  `       ``(K ``=``=` `N ``and` `oddCount ``%` `2` `=``=` `0``)): ` `        ``return` `False` `    ``else``: ` `        ``return` `True` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5``, ``8` `] ` `    ``K ``=` `5` `    ``N ``=` `len``(arr) ` ` `  `    ``if` `(isPossible(arr, N, K)): ` `        ``print``(``"Possible"``) ` `    ``else``: ` `        ``print``(``"Not Possible"``) ` ` `  `# This code is contributed by mohit kumar 29     `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function returns true if  ` `// it is possible to have  ` `// odd sum  ` `static` `bool` `isPossible(``int` `[]arr,  ` `                       ``int` `N, ``int` `K)  ` `{  ` `    ``int` `oddCount = 0, evenCount = 0;  ` ` `  `    ``// Counting number of odd  ` `    ``// and even elements  ` `    ``for``(``int` `i = 0; i < N; i++) ` `    ``{  ` `       ``if` `(arr[i] % 2 == 0) ` `       ``{  ` `           ``evenCount++; ` `       ``}  ` `       ``else` `       ``{ ` `           ``oddCount++; ` `       ``} ` `    ``}  ` `    ``if` `(evenCount == N ||  ` `       ``(oddCount == N && K % 2 == 0) ||  ` `       ``(K == N && oddCount % 2 == 0)) ` `    ``{  ` `        ``return` `false``; ` `    ``}  ` `    ``else` `    ``{ ` `        ``return` `true``; ` `    ``}  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main (``string``[] args)  ` `{  ` `    ``int` `[]arr = { 1, 2, 3, 4, 5, 8 };  ` `    ``int` `K = 5;  ` `    ``int` `N = arr.Length;  ` ` `  `    ``if` `(isPossible(arr, N, K)) ` `    ``{ ` `        ``Console.WriteLine(``"Possible"``); ` `    ``}  ` `    ``else` `    ``{ ` `        ``Console.WriteLine(``"Not Possible"``); ` `    ``}  ` `}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```Possible
```

Time complexity: O(N)

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Improved By : mohit kumar 29, AnkitRai01

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