3D Surface Area

Madison is a little girl who is fond of toys. Her friend Mason works in a toy manufacturing factory. Mason has a 2D board A of size with H rows and W columns. The board is divided into cells of size with each cell indicated by its coordinate (i,j). The cell (i,j) has an integer A_ij written on it. To create the toy Mason stacks A_ij number of cubes of size on the cell (i, j).

Given the description of the board showing the values of A_ij and that the price of the toy is equal to the 3d surface area find the price of the toy.

Input Format

The first line contains two space-separated integers H and W the height and the width of the board respectively.

The next H lines contains W space separated integers. The j^th integer in i^th line denotes A_ij.

Constraints

• 1 <= H, W <= 100
• 

Output Format

Print the required answer, i.e the price of the toy, in one line.

Sample Input 0

1
2

1 1
1

Sample Output 0

1

6

Explanation 0

The surface area of cube is 6.

Sample Input 1

1
2
3
4

3 3
1 3 4
2 2 3
1 2 4

Sample Output 1

1

60

Explanation 1

The object is rotated so the front row matches column 1 of the input, heights 1, 2, and 1.

The front face is 1 + 2 + 1 = 4 units in area.
The top is 3 units.
The sides are 4 units.
None of the rear faces are exposed.
The underside is 3 units.
The front row contributes 4 + 3 + 4 + 3 = 14 units to the surface area.

Solution

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/*
 * Complete the 'surfaceArea' function below.
 *
 * The function is expected to return an INTEGER.
 * The function accepts 2D_INTEGER_ARRAY A as parameter.
 */

function surfaceArea(A) {
  // Write your code here
  let count = 0;

  for (let i = 0, itotal = A.length; i < itotal; i++) {
    for (let j = 0, jtotal = A[0].length; j < jtotal; j++) {
      count +=
        [
          (A[i - 1] && A[i - 1][j]) || 0,
          (A[i + 1] && A[i + 1][j]) || 0,
          A[i][j - 1] || 0,
          A[i][j + 1] || 0
        ].reduce((target, item) => target + Math.max(0, A[i][j] - item), 0) + 2;
    }
  }

  return count;
}