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Check if K distinct array elements form an odd sum

  • Last Updated : 09 Jun, 2021

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 an odd number of 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++




// 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;
}

Java




// 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

Python3




# Python3 program for
# the above approach
 
# Function to return if
# odd sum is possible or not
def oddSum(A, N, K):
 
    # Stores distinct odd elements
    Odd = set([])
    # Stores distinct even elements
    Even = set([])
  
    # Iterating through given array
    for i in range (N):
  
        # 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 (len(Odd) >= K):
        return True
  
    flag = False
  
    # Check for all odd frequencies
    # of odd elements whether
    # sufficient even numbers
    # are present or not
    for i in range (1, K, 2):
  
        # Count of even numbers
        # required
        needed = K - i
  
        # If required even numbers
        # are present in the array
        if (needed <= len(Even)):
            return True
  
    return flag
  
# Driver Program
if __name__ == "__main__":
 
    K = 5
    A = [12, 1, 7,
         7, 26, 18]
    N = 3
  
    if (oddSum(A, N, K)):
        print ("YES")
    else:
        print("NO")
  
# This code is contributed by Chitranayal

C#




// 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

Javascript




<script>
 
// JavaScript program for the above approach
 
    // Function to return if
   // odd sum is possible or not
    function oddSum(A,N,K)
    {
        // Stores distinct odd elements
    let Odd = new Set();
      
    // Stores distinct even elements
    let Even = new Set();
  
    // Iterating through given array
    for(let 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;
  
    let flag = false;
  
    // Check for all odd frequencies
    // of odd elements whether
    // sufficient even numbers
    // are present or not
    for(let i = 1; i < K; i += 2)
    {
          
        // Count of even numbers
        // required
        let needed = K - i;
              
        // If required even numbers
        // are present in the array
        if (needed <= Even.size)
        {
            return true;
        }
    }
    return flag;
    }
     
    // Driver code
    let K = 5;
    let A=[12, 1, 7, 7, 26, 18];
    let N = 3;
    if (oddSum(A, N, K))
        document.write("YES");
    else
        document.write("NO");
 
 
// This code is contributed by avanitrachhadiya2155
 
</script>
Output: 
NO

 

Time complexity: O(N)




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