Projection Area of 3D Shapes

Similar Problems:

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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

- Solution:

// Blog link: https://code.dennyzhang.com/projection-area-of-3d-shapes // Basic Ideas: math // Notice: the matrix is a square, instead of a rectangle // Complexity: Time O(n*n), Space O(1) func projectionArea(grid [][]int) int { res := 0 n := len(grid) for i:=0; i<n; i++ { max_row, max_col := -1<<31, -1<<31 for j:=0; j<n; j++ { if grid[i][j] > max_row { max_row = grid[i][j] } if grid[j][i] > max_col { max_col = grid[j][i] } if grid[i][j] != 0 { res++ } } res += max_row + max_col } return res }

// Blog link: https://code.dennyzhang.com/projection-area-of-3d-shapes // Basic Ideas: math // Complexity: Time O(n*n), Space O(1) func projectionArea(grid [][]int) int { res := 0 for _, row := range grid { max := -1<<31 for _, v := range row { if v > max { max = v } if v != 0 { res++ } } res += max } for j:=0; j<len(grid[0]); j++ { max := -1<<31 for i:=0; i<len(grid); i++ { if grid[i][j] > max { max = grid[i][j] } } res += max } return res }