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Formatted question description: https://leetcode.ca/all/892.html

892. Surface Area of 3D Shapes (Easy)

On a N * N grid, we place some 1 * 1 * 1 cubes.

Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell (i, j).

Return the total surface area of the resulting shapes.

 

Example 1:

Input: [[2]]
Output: 10

Example 2:

Input: [[1,2],[3,4]]
Output: 34

Example 3:

Input: [[1,0],[0,2]]
Output: 16

Example 4:

Input: [[1,1,1],[1,0,1],[1,1,1]]
Output: 32

Example 5:

Input: [[2,2,2],[2,1,2],[2,2,2]]
Output: 46

 

Note:

• 1 <= N <= 50
• 0 <= grid[i][j] <= 50

Solution 1.

// OJ: https://leetcode.com/problems/surface-area-of-3d-shapes/
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
    int surfaceArea(vector<vector<int>>& grid) {
        int N = grid.size(), area = 0;
        for (int i = 0; i < N; ++i) {
            int prev = 0;
            for (int j = 0; j < N; ++j) {
                if (grid[i][j]) area += 2;
                area += abs(grid[i][j] - prev);
                prev = grid[i][j];
            }
            area += prev;
        }
        for (int i = 0; i < N; ++i) {
            int prev = 0;
            for (int j = 0; j < N; ++j) {
                area += abs(grid[j][i] - prev);
                prev = grid[j][i];
            }
            area += prev;
        }
        return area;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/surface-area-of-3d-shapes/
// Time: O(N^2)
// Space: O(1)
class Solution {
private:
    int dirs[4][2] = { {0,1}, {-1,0}, {0,-1}, {1,0} };
public:
    int surfaceArea(vector<vector<int>>& grid) {
        int N = grid.size(), area = 0;
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < N; ++j) {
                if (!grid[i][j]) continue;
                area += 2;
                for (auto &dir : dirs) {
                    int x = i + dir[0], y = j + dir[1];
                    int h = x < 0 || x >= N || y < 0 || y >= N ? 0 : grid[x][y];
                    area += max(grid[i][j] - h, 0);
                }
            }
        }
        return area;
    }
};

Java

• Java
• C++
• Python

• class Solution {
      public int surfaceArea(int[][] grid) {
          int area = 0;
          int[][] directions = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
          int rows = grid.length, columns = grid[0].length;
          for (int i = 0; i < rows; i++) {
              for (int j = 0; j < columns; j++) {
                  int cell = grid[i][j];
                  int curArea = cell == 0 ? 0 : cell * 4 + 2;
                  for (int[] direction : directions) {
                      int newRow = i + direction[0], newColumn = j + direction[1];
                      if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns) {
                          int adjacent = grid[newRow][newColumn];
                          int overlapArea = Math.min(cell, adjacent);
                          curArea -= overlapArea;
                      }
                  }
                  area += curArea;
              }
          }
          return area;
      }
  }
  

• // OJ: https://leetcode.com/problems/surface-area-of-3d-shapes/
  // Time: O(N^2)
  // Space: O(1)
  class Solution {
  public:
      int surfaceArea(vector<vector<int>>& grid) {
          int N = grid.size(), area = 0;
          for (int i = 0; i < N; ++i) {
              int prev = 0;
              for (int j = 0; j < N; ++j) {
                  if (grid[i][j]) area += 2;
                  area += abs(grid[i][j] - prev);
                  prev = grid[i][j];
              }
              area += prev;
          }
          for (int i = 0; i < N; ++i) {
              int prev = 0;
              for (int j = 0; j < N; ++j) {
                  area += abs(grid[j][i] - prev);
                  prev = grid[j][i];
              }
              area += prev;
          }
          return area;
      }
  };
  

• class Solution:
      def surfaceArea(self, grid: List[List[int]]) -> int:
          ans = 0
          for i, row in enumerate(grid):
              for j, v in enumerate(row):
                  if v:
                      ans += 2 + v * 4
                      if i:
                          ans -= min(v, grid[i - 1][j]) * 2
                      if j:
                          ans -= min(v, grid[i][j - 1]) * 2
          return ans
  
  ############
  
  class Solution(object):
      def surfaceArea(self, grid):
          """
          :type grid: List[List[int]]
          :rtype: int
          """
          area = 0
          n = len(grid)
          for i in range(n):
              for j in range(n):
                  if grid[i][j]: area += grid[i][j] * 4 + 2
                  if i: area -= min(grid[i][j], grid[i-1][j]) * 2
                  if j: area -= min(grid[i][j], grid[i][j-1]) * 2
          return area
  

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