883

883. Projection Area of 3D Shapes (Easy)

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

Related Topics:
Math

Solution 1.

// OJ: https://leetcode.com/problems/projection-area-of-3d-shapes/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(1)
class Solution {
public:
    int projectionArea(vector<vector<int>>& grid) {
        int cnt = 0;
        for (int N = grid.size(), i = 0; i < N; ++i) {
            int max1 = 0, max2 = 0;
            for (int j = 0; j < N; ++j) {
                if (grid[i][j]) ++cnt;
                max1 = max(max1, grid[i][j]);
                max2 = max(max2, grid[j][i]);
            }
            cnt += max1 + max2;
        }
        return cnt;
    }
};