Question

883. Projection Area of 3D Shapes

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Description

You are given an n x n grid where we place some 1 x 1 x 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 the cell (i, j).

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

Example 2:

Input: grid = [[2]]
Output: 5

Example 3:

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

 

Constraints:

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

Solutions

Solution 1: Mathematics

We can calculate the area of the three projections separately.

• Projection area on the xy plane: Each non-zero value will be projected onto the xy plane, so the projection area on the xy plane is the count of non-zero values.
• Projection area on the yz plane: The maximum value in each row.
• Projection area on the zx plane: The maximum value in each column.

Finally, add up the three areas.

The time complexity is $O(n^2)$, where $n$ is the side length of the grid grid. The space complexity is $O(1)$.

class Solution:
  def projectionArea(self, grid: List[List[int]]) -> int:
    xy = sum(v > 0 for row in grid for v in row)
    yz = sum(max(row) for row in grid)
    zx = sum(max(col) for col in zip(*grid))
    return xy + yz + zx

class Solution {
  public int projectionArea(int[][] grid) {
    int xy = 0, yz = 0, zx = 0;
    for (int i = 0, n = grid.length; i < n; ++i) {
      int maxYz = 0;
      int maxZx = 0;
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] > 0) {
          ++xy;
        }
        maxYz = Math.max(maxYz, grid[i][j]);
        maxZx = Math.max(maxZx, grid[j][i]);
      }
      yz += maxYz;
      zx += maxZx;
    }
    return xy + yz + zx;
  }
}

class Solution {
public:
  int projectionArea(vector<vector<int>>& grid) {
    int xy = 0, yz = 0, zx = 0;
    for (int i = 0, n = grid.size(); i < n; ++i) {
      int maxYz = 0, maxZx = 0;
      for (int j = 0; j < n; ++j) {
        xy += grid[i][j] > 0;
        maxYz = max(maxYz, grid[i][j]);
        maxZx = max(maxZx, grid[j][i]);
      }
      yz += maxYz;
      zx += maxZx;
    }
    return xy + yz + zx;
  }
};

func projectionArea(grid [][]int) int {
  xy, yz, zx := 0, 0, 0
  for i, row := range grid {
    maxYz, maxZx := 0, 0
    for j, v := range row {
      if v > 0 {
        xy++
      }
      maxYz = max(maxYz, v)
      maxZx = max(maxZx, grid[j][i])
    }
    yz += maxYz
    zx += maxZx
  }
  return xy + yz + zx
}

function projectionArea(grid: number[][]): number {
  const xy: number = grid.flat().filter(v => v > 0).length;
  const yz: number = grid.reduce((acc, row) => acc + Math.max(...row), 0);
  const zx: number = grid[0]
    .map((_, i) => Math.max(...grid.map(row => row[i])))
    .reduce((acc, val) => acc + val, 0);
  return xy + yz + zx;
}

impl Solution {
  pub fn projection_area(grid: Vec<Vec<i32>>) -> i32 {
    let xy: i32 = grid
      .iter()
      .map(
        |row|
          row
            .iter()
            .filter(|&&v| v > 0)
            .count() as i32
      )
      .sum();
    let yz: i32 = grid
      .iter()
      .map(|row| *row.iter().max().unwrap_or(&0))
      .sum();
    let zx: i32 = (0..grid[0].len())
      .map(|i|
        grid
          .iter()
          .map(|row| row[i])
          .max()
          .unwrap_or(0)
      )
      .sum();
    xy + yz + zx
  }
}