surface Archives - Huahua's Tech Road https://zxi.mytechroad.com/blog/tag/surface/ Thu, 30 Aug 2018 16:32:26 +0000 en hourly 1 https://wordpress.org/?v=6.0.8 https://zxi.mytechroad.com/blog/wp-content/uploads/2017/09/cropped-photo-32x32.jpg surface Archives - Huahua's Tech Road https://zxi.mytechroad.com/blog/tag/surface/ 32 32 花花酱 LeetCode 892. Surface Area of 3D Shapes https://zxi.mytechroad.com/blog/geometry/leetcode-892-surface-area-of-3d-shapes/ https://zxi.mytechroad.com/blog/geometry/leetcode-892-surface-area-of-3d-shapes/#respond Sun, 26 Aug 2018 15:25:33 +0000 https://zxi.mytechroad.com/blog/?p=3701 Problem 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…

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]]> Problem

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: Geometry

3D version of 花花酱 LeetCode 463. Island Perimeter

Ans = total faces – hidden faces.

each pile with height h has the surface area of 2 (top/bottom) + 4*h (sides)

\(ans = ans + 2 + 4 * h\)

if a cube has a neighbour, reduce the total surface area by 1

For each pile, we check 4 neighbours, the number of total hidden faces of this pile is sum(min(h, neighbour’s h)).

\(ans = ans – \Sigma min(h, n.h)\)

Time complexity: O(mn)

Space complexity: O(1)

C++

// Author: Huahua
// Running time: 4 ms
class Solution {
public:
  int surfaceArea(vector<vector<int>>& grid) {
    static const vector<int> dirs{0, -1, 0, 1, 0};
    int m = grid.size();
    int n = grid[0].size();
    int ans = 0;
    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j) {
        int h = grid[i][j];
        if (h == 0) continue;
        ans += 2 + 4 * h;        
        for (int k = 0; k < 4; k++) {
          int tx = j + dirs[k];
          int ty = i + dirs[k + 1];
          if (tx < 0 || tx >= n || ty < 0 || ty >= m) continue;
          int th = grid[ty][tx];          
          ans -= (th <= 0 ? 0 : min(h, th));
        }
      }
    return ans;
  }
};

C++ (opt)

Since the neighbor relationship is symmetric, we can only consider the top and left neighbors and double the hidden faces.

// Author: Huahua
// Running time: 4 ms
class Solution {
public:
  int surfaceArea(vector<vector<int>>& grid) {    
    int m = grid.size();
    int n = grid[0].size();
    int ans = 0;
    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j) {
        int h = grid[i][j];
        if (h == 0) continue;
        ans += 2 + 4 * h;
        if (i) ans -= 2 * min(h, grid[i - 1][j]);
        if (j) ans -= 2 * min(h, grid[i][j - 1]);        
      }
    return ans;
  }
};

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