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 <= 500 <= 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++
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// Author: Huahua // Running time : 4 ms class Solution { public: int projectionArea(vector<vector<int>>& 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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