Problem:
Score : 300 points

Problem Statement

There are N candles placed on a number line.
The i-th candle from the left is placed on coordinate x_i.
Here, x_1 < x_2 < ... < x_N holds.

Initially, no candles are burning.
Snuke decides to light K of the N candles.

Now, he is at coordinate 0.
He can move left and right along the line with speed 1.
He can also light a candle when he is at the same position as the candle, in negligible time.

Find the minimum time required to light K candles.

Constraints

1 \leq N \leq 10^5

1 \leq K \leq N

x_i is an integer.

|x_i| \leq 10^8

x_1 < x_2 < ... < x_N

Input

Input is given from Standard Input in the following format:

N K
x_1 x_2 ... x_N

Output

Print the minimum time required to light K candles.

Sample Input 1

5 3
-30 -10 10 20 50

Sample Output 1

40

He should move and light candles as follows:

Move from coordinate 0 to -10.

Light the second candle from the left.

Move from coordinate -10 to 10.

Light the third candle from the left.

Move from coordinate 10 to 20.

Light the fourth candle from the left.

Sample Input 2

3 2
10 20 30

Sample Output 2

20

Sample Input 3

1 1
0

Sample Output 3

0

There may be a candle placed at coordinate 0.

Sample Input 4

8 5
-9 -7 -4 -3 1 2 3 4

Sample Output 4

10

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.Arrays;
import java.util.Scanner;


public class Main {
	static int N,K;
	static int[] T;
	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		 N = sc.nextInt();
		 K = sc.nextInt();
		 T = new int[N+1];

		 int p = 0;
		 int q = 0 ;
		 for(int i = 0 ; i < N ; i++){
			 T[i]= sc.nextInt();
			 if(T[i]<0){
				 p++;
			 }else if(T[i]>0){
				 q++;
			 }
		 }

		 sc.close();

		 if(N-p-q == 0){
			 T[N] = 0;
			 Arrays.sort(T);
		 }

		 //System.out.println("負数 : "+p);
		 //System.out.println("正数 : "+q);

		 //ゼロの項番
		 int zero = p;



		 int ans = 100000001 ;
		 for(int i = 0 ; i < K ; i++){
			 if(zero-i>=0 && zero+K-i<=N){
				 int pp = Math.abs(T[zero-i])*2 + Math.abs(T[zero+K-i]);
				 //System.out.println("plus pp : "+pp);
				 if(pp < ans){
					 ans = pp;
				 }
			 }
			 else{
				 //System.out.println("plus continue");
				 continue;
			 }
		}
		 for(int i = 0 ; i < K ; i++){

			 if(zero+i<=N && zero-K+i>=0){
				 int pp = Math.abs(T[zero+i])*2 + Math.abs(T[zero-K+i]);
				 //System.out.println("minus pp : "+pp);
				 if(pp < ans){
					 ans = pp;
				 }
			 }
			 else{
				 //System.out.println("minus continue");
				 continue;
			 }
		}
		 System.out.println(ans);


 	}
}