Check if K distinct array elements form an odd sum

Given an array A[] of size N, the task is to check if it is possible to get odd sum using K distinct elements from the array.

Examples:

Input: N = 4, K = 2, A = {10, 19, 14, 14}
Output: YES
Explanation:
19 + 14 = 33, which is odd

Input: N = 3, K = 3, A = {101, 102, 103}
Output: NO
Explanation:
101 + 102 + 103 = 306, which is even

Approach:



  • Let us look at some basic maths properties:
    EVEN + EVEN = EVEN
    ODD  + ODD  = EVEN
    EVEN + ODD  = ODD
    
  • From the above properties, we can conclude that when odd numbe rof odd integers are added to any number of even integers, the result will always be odd.

    Examples:

    • 3 + 2 + 6 = 11 (odd)
    • 1 + 3 + 7 + 4 = 15 (odd)
  • Hence, follow the steps below to solve the problem:

    • Count distinct odd and even elements present in the array.
    • If K or more odd elements are present in the array, print “Yes”.
    • Otherwise, for every odd count of odd numbers, check if sufficient even numbers are present in the array or not.
    • If any such case occurs, print “Yes”. Otherwise, print “No”.

    Below is the implementation of the above approach:

    C++

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    // C++ program for above approach 
      
    #include <bits/stdc++.h> 
    using namespace std; 
      
    // Function to return if 
    // odd sum is possible or not 
    bool oddSum(vector<int>& A, 
                int N, int K) 
      
        // Stores distinct odd elements 
        set<int> Odd; 
        // Stores distinct even elements 
        set<int> Even; 
      
        // Iterating through given array 
        for (int i = 0; i < N; i++) { 
      
            // If element is even 
            if (A[i] % 2 == 0) { 
                Even.insert(A[i]); 
            
            // If element is odd 
            else
                Odd.insert(A[i]); 
            
        
      
        // If atleast K elements 
        // in the array are odd 
        if (Odd.size() >= K) 
            return true
      
        bool flag = false
      
        // Check for all odd frequencies 
        // of odd elements whether 
        // sufficient even numbers 
        // are present or not 
        for (int i = 1; i < K; i += 2) { 
      
            // Count of even numbers 
            // required 
            int needed = K - i; 
      
            // If required even numbers 
            // are present in the array 
            if (needed <= Even.size()) { 
      
                return true
            
        
      
        return flag; 
      
    // Driver Program 
    int main() 
        int K = 5; 
        vector<int> A = { 12, 1, 7, 7, 26, 18 }; 
        int N = 3; 
      
        if (oddSum(A, N, K)) 
            cout << "YES"
        else
            cout << "NO"
      
        return 0; 

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    Java

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    // Java program for the above approach 
    import java.util.*; 
      
    class GFG{ 
      
    // Function to return if 
    // odd sum is possible or not 
    static boolean oddSum(int []A, 
                          int N, int K) 
          
        // Stores distinct odd elements 
        HashSet<Integer> Odd = new HashSet<Integer>(); 
          
        // Stores distinct even elements 
        HashSet<Integer> Even = new HashSet<Integer>(); 
      
        // Iterating through given array 
        for(int i = 0; i < N; i++) 
        
              
            // If element is even 
            if (A[i] % 2 == 0
            
                Even.add(A[i]); 
            
                  
            // If element is odd 
            else
            
                Odd.add(A[i]); 
            
        
      
        // If atleast K elements 
        // in the array are odd 
        if (Odd.size() >= K) 
            return true
      
        boolean flag = false
      
        // Check for all odd frequencies 
        // of odd elements whether 
        // sufficient even numbers 
        // are present or not 
        for(int i = 1; i < K; i += 2
        
              
            // Count of even numbers 
            // required 
            int needed = K - i; 
                  
            // If required even numbers 
            // are present in the array 
            if (needed <= Even.size()) 
            
                return true
            
        
        return flag; 
      
    // Driver code 
    public static void main(String[] args) 
        int K = 5
        int []A = { 12, 1, 7, 7, 26, 18 }; 
        int N = 3
      
        if (oddSum(A, N, K)) 
            System.out.print("YES"); 
        else
            System.out.print("NO"); 
      
    // This code is contributed by PrinciRaj1992 

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    C#

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    // C# program for the above approach
    using System;
    using System.Collections.Generic;
      
    class GFG{
      
    // Function to return if
    // odd sum is possible or not
    static bool oddSum(int []A, int N, int K)
    {
          
        // Stores distinct odd elements
        HashSet<int> Odd = new HashSet<int>();
          
        // Stores distinct even elements
        HashSet<int> Even = new HashSet<int>();
      
        // Iterating through given array
        for(int i = 0; i < N; i++)
        {
              
            // If element is even 
            if (A[i] % 2 == 0)  
            
                Even.Add(A[i]); 
            
              
            // If element is odd 
            else
            
                Odd.Add(A[i]); 
            
        }
          
        // If atleast K elements 
        // in the array are odd 
        if (Odd.Count >= K) 
            return true
      
        bool flag = false;
      
        // Check for all odd frequencies
        // of odd elements whether
        // sufficient even numbers
        // are present or not
        for(int i = 1; i < K; i += 2)
        {
              
            // Count of even numbers
            // required
            int needed = K - i;
                  
            // If required even numbers
            // are present in the array
            if (needed <= Even.Count)
            {
                return true;
            }
        }
        return flag;
    }
      
    // Driver code
    public static void Main(String[] args)
    {
        int K = 5;
        int []A = { 12, 1, 7, 7, 26, 18 };
        int N = 3;
      
        if (oddSum(A, N, K))
            Console.Write("YES");
        else
            Console.Write("NO");
    }
    }
      
    // This code is contributed by PrinciRaj1992

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    Output:

    NO
    

    Time complexity: O(N)

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