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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.

• Java
• C++
• Python
• Go

• 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;
      }
  }
  
  ############
  
  class Solution {
      public int surfaceArea(int[][] grid) {
          int n = grid.length;
          int ans = 0;
          for (int i = 0; i < n; ++i) {
              for (int j = 0; j < n; ++j) {
                  if (grid[i][j] > 0) {
                      ans += 2 + grid[i][j] * 4;
                      if (i > 0) {
                          ans -= Math.min(grid[i][j], grid[i - 1][j]) * 2;
                      }
                      if (j > 0) {
                          ans -= Math.min(grid[i][j], grid[i][j - 1]) * 2;
                      }
                  }
              }
          }
          return ans;
      }
  }
  

• // 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
  

• func surfaceArea(grid [][]int) int {
  	ans := 0
  	for i, row := range grid {
  		for j, v := range row {
  			if v > 0 {
  				ans += 2 + v*4
  				if i > 0 {
  					ans -= min(v, grid[i-1][j]) * 2
  				}
  				if j > 0 {
  					ans -= min(v, grid[i][j-1]) * 2
  				}
  			}
  		}
  	}
  	return ans
  }
  
  func min(a, b int) int {
  	if a < b {
  		return a
  	}
  	return b
  }
  

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