花花酱 LeetCode 887. Projection Area of 3D Shapes

zxi — Sun, 05 Aug 2018 03:06:30 +0000

Problem

On a N * N grid, we place some 1 * 1 * 1 cubes that are axis-aligned with the x, y, and z axes.

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

Now we view the projection of these cubes onto the xy, yz, and zx planes.

A projection is like a shadow, that maps our 3 dimensional figure to a 2 dimensional plane.

Here, we are viewing the “shadow” when looking at the cubes from the top, the front, and the side.

Return the total area of all three projections.

Example 1:

Input: [[2]]
Output: 5

Example 2:

Input: [[1,2],[3,4]]
Output: 17
Explanation: 
Here are the three projections ("shadows") of the shape made with each axis-aligned plane.

Example 3:

Input: [[1,0],[0,2]]
Output: 8

Example 4:

Input: [[1,1,1],[1,0,1],[1,1,1]]
Output: 14

Example 5:

Input: [[2,2,2],[2,1,2],[2,2,2]]
Output: 21

Note:

1 <= grid.length = grid[0].length <= 50
0 <= grid[i][j] <= 50

Solution: Brute Force

Sum of max heights for each cols / rows + # of non-zero-height bars.

Time complexity: O(mn)

Space complexity: O(1)

C++

// Author: Huahua
// Running time : 4 ms
class Solution {
public:
  int projectionArea(vector>& grid) {
    int n = grid.size();    
    int m = grid[0].size();
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      int h = 0;
      for (int j = 0; j < m; ++j) {
        h = max(h, grid[i][j]);
        if (grid[i][j] != 0) ++ans;
      }
      ans += h;      
    }
    for (int j = 0; j < m; ++j) {
      int h = 0;
      for (int i = 0; i < n; ++i)
        h = max(h, grid[i][j]);
      ans += h;      
    }
    return ans;
  }
};

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花花酱 LeetCode 887. Projection Area of 3D Shapes

Problem

Solution: Brute Force