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 C++14 (GCC 5.4.1) (judge verdict: Accepted)


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

long long X[ 100010 ];

int main()
{
int i;
int j;
int a;
int N;
int K;
long long iMin;
bool b;
int P;
int M;
int p;
int m;

	ios::sync_with_stdio( false );
	cin.tie( 0 );

	cin >> N;
	cin >> K;

	for( i = 0; i < N; i++ ) cin >> X[ i ];

	sort( X, X + N );

	b = false;
	P = - 1;
	M = N - 1;
	for( i = 0; i < N; i++ )
	{
		if( X[ i ] == 0 )
		{
			b = true;
			P = i;
			M = i;

			break;
		}
		else if( X[ i ] > 0 )
		{
			P = i;
			M = i - 1;

			break;
		}
	}

	if( M < 0 )
	{
		cout << X[ K - 1 ] << endl;
		return 0;
	}
	else if( P < 0 )
	{
		cout << - X[ N - K ] << endl;
		return 0;
	}

	iMin = 1000000000;
	for( i = 0; i < N; i++ )
	{
		p = i - P + 1;
		m = M - i + 1;
		if( m > K ) continue;
		if( p > K ) break;

		if( m == K )
		{
			iMin = min( iMin, - X[ i ] );
		}
		else if( p == K )
		{
			iMin = min( iMin, X[ i ] );
		}
		else if( i <= M )
		{
			a = K - m + P;
			if( ! b ) a--;
			if( a < 0 ) continue;
			if( a >= N ) continue;

			iMin = min( iMin, - 2 * X[ i ] + X[ a ] );
		}
		else if( i >= P )
		{
			a = M - ( K - p );
			if( ! b ) a++;
			if( a < 0 ) continue;
			if( a >= N ) continue;

			iMin = min( iMin, 2 * X[ i ] - X[ a ] );
		}
	}

	cout << iMin << endl;

	return 0;
}