## Solution

#### Approach 1: Square by Square

Intuition

Let's try to count the surface area contributed by v = grid[i][j].

When v > 0, the top and bottom surface contributes an area of 2.

Then, for each side (west side, north side, east side, south side) of the column at grid[i][j], the neighboring cell with value nv means our square contributes max(v - nv, 0).

For example, for grid = [[1, 5]], the contribution at grid[0][1] is 2 + 5 + 5 + 5 + 4. The 2 comes from the top and bottom side, the 5 comes from the north, east, and south side; and the 4 comes from the west side, of which 1 unit is covered by the adjacent column.

Algorithm

For each v = grid[r][c] > 0, count ans += 2, plus ans += max(v - nv, 0) for each neighboring value nv adjacent to grid[r][c].

Complexity Analysis

• Time Complexity: , where is the number of rows (and columns) in the grid.

• Space Complexity: .

Analysis written by: @awice.