LeetCode: Projection Area of 3D Shapes

Posted on August 23, 2018July 26, 2020 by braindenny

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:

// 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^2), 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
}

// https://code.dennyzhang.com/projection-area-of-3d-shapes
// Basic Ideas: math
// Complexity: Time O(n^2), 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
}

Post Views: 4