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.*;

class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int K = sc.nextInt();
        long[] x = new long[N];
        for(int i=0;i<N;i++) x[i] = sc.nextLong();
        long[] dis = new long[N];
        for(int i=0;i<N;i++) dis[i]=Math.abs(x[i]);
        if(x[0]>=0){
            System.out.println(x[K-1]);
        }else if(x[N-1]<=0){
            System.out.println(Math.abs(x[N-K]));
        }
        else{
            long ans1 = Long.MAX_VALUE;
            for(int i=0;i<N-K;i++){
                long cur = dis[i]+x[i+K-1]-x[i];
                ans1 = Math.min(ans1,cur);
            }
            long ans2 = Long.MAX_VALUE;
            for(int i=N-1;i>=K-1;i--){
                long cur = dis[i]+x[i]-x[i-K+1];
                ans2 = Math.min(ans2,cur);
            }
            System.out.println(Math.min(ans1,ans2));
        }
    }
}