The Wars to Come | Editorial by shivamjaiswal64

In this approach, We have to find the minimum number of  Wildlings.
We will follow these steps:
* Input the number of of Wildlings.
* Take and store the strength in Array ar[] of each wildling.
* Input the minimum strength that is demanded.
* Sort the array 
* Traverse the reverse array and add the strength in sum (till sum <= min).
* During traversing , count the number and store in variable c .
* Print the value of c for O/P of minimum number of  Wildlings..

Java8 code:

```
import java.util.Scanner;
import java.util.Arrays;
class Apr2020{
    public static void main(String args[] ) throws Exception {
        Scanner s = new Scanner(System.in);
        int t=s.nextInt();
        int ar[] = new int[t];
        for(int i=0;i<t;i++)
            ar[i] = s.nextInt();
        Arrays.sort(ar);
        int min = s.nextInt();
        int sum =0,c=0;
        for(int i=t-1;i>=0;i--){
            if(sum <= min){
                sum += ar[i];
                c++;
            }
            else
                break;
        }
        System.out.print(c); 
    }
}

```
Editorial By: [Shivam Jaiswal](https://www.stopstalk.com/user/profile/shivamjaiswal64)

The Wars to Come | Editorial by shivamjaiswal64

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