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

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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    

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C#

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// 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

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Output:

Possible

Time complexity: O(N)

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