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[] = {2, 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 

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 <bits/stdc++.h>
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[0]);

    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

Javascript

<script>
// Javascript implementation of the above approach

// Function returns true if
// it is possible to have
// odd sum
function isPossible(arr, N, K)
{
    let oddCount = 0, evenCount = 0;

    // Counting number of odd
    // and even elements
    for(let 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

    let arr = [ 1, 2, 3, 4, 5, 8 ];
    let K = 5;
    let N = arr.length;

    if (isPossible(arr, N, K))
    {
        document.write("Possible");
    }
    else
    {
        document.write("Not Possible");
    }

// This code is contributed by target_2.
</script>

Output

Possible

Time complexity: O(N), where N represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.