Odd Selection | Editorial by Dipali

Odd Selection | Editorial by Dipali_Kumbhar

**Basic condition:** The sum of numbers is odd if and only if there are odd number of odd numbers.
**Explanation:** 
1. if(count_odd >= x) then it is possible to get odd sum by using given numbers.
2. else  check whether count_odd is even or odd.
3. Refer below code for more clarification. 
**Note: ** There might be different solutions. Below is one of them.
**Complexity:**  O(n)

**Code: **
```
/* Problem : Odd Selection */
/* Author : Dipali Kumbhar */

#include<bits/stdc++.h>
using namespace std;

//define macros required
#define ll long long
#define endl "\n"
#define vll vector<ll>
#define pb push_back

//use fast input-output methods in c++
void fastio()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}

int main()
{
    fastio();
    ll t;
    cin>>t;
    while(t--)
    {
        ll n,x;
        cin>>n>>x;
        vll v;                  //dynamic array declaration(vector)
        
        ll count_odd=0;          //to store number of odd no.
        ll count_even=0;        //to store number of even no.
        for(ll i=0;i<n;i++)
        {
            ll y;
            cin>>y;
            v.pb(y);
            if(y%2==0)
                count_even++;
            else
                count_odd++;
            
        }
        if(count_odd == 0)      //if odd numbers are not present in a given array
            cout<<"No"<<endl;
        
        //if count_odd is equal to x but x is odd
        //then we can choose any x numbers from array
        else if(x<=count_odd && x%2 !=0)        
        {
            cout<<"Yes"<<endl;                  
        }
        
        //if count_odd is equal to x but x is even
        //then we have to choose x-1 odd numbers 
        //and one even number
        else if(x<=count_odd && x%2 == 0)
        {
            if(count_even>=1)
                cout<<"Yes"<<endl;
            else
                cout<<"No"<<endl;
        }
        
        //if count_odd is less than x
        else
        {
            //if count_odd is odd
            //then we can choose all odd numbers
            //and x-count_odd even numbers
            if(count_odd%2 !=0)
            {
                x-=count_odd;
                if(count_even>=x)
                    cout<<"Yes"<<endl;
                else
                    cout<<"No"<<endl;
            }
            
            //if count_odd is even
            //then we can choose all odd numbers except one
            //and x-count_odd even numbers
            else
            {
                x-=count_odd-1;
                if(count_even>=x)
                    cout<<"Yes"<<endl;
                else
                    cout<<"No"<<endl;
            }
        }
    }
    
    
    return 0;
}
```

Odd Selection | Editorial by Dipali_Kumbhar

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