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883. Projection Area of 3D Shapes

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

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  };
  

• 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
  
  

• 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 n = grid.length;
      let res = grid.reduce((r, v) => r + v.reduce((r, v) => r + (v === 0 ? 0 : 1), 0), 0);
      for (let i = 0; i < n; i++) {
          let xMax = 0;
          let yMax = 0;
          for (let j = 0; j < n; j++) {
              xMax = Math.max(xMax, grid[i][j]);
              yMax = Math.max(yMax, grid[j][i]);
          }
          res += xMax + yMax;
      }
      return res;
  }
  
  

• impl Solution {
      pub fn projection_area(grid: Vec<Vec<i32>>) -> i32 {
          let n = grid.len();
          let mut res = 0;
          let mut x_max = vec![0; n];
          let mut y_max = vec![0; n];
          for i in 0..n {
              for j in 0..n {
                  let val = grid[i][j];
                  if val == 0 {
                      continue;
                  }
                  res += 1;
                  x_max[i] = x_max[i].max(val);
                  y_max[j] = y_max[j].max(val);
              }
          }
          res + y_max.iter().sum::<i32>() + x_max.iter().sum::<i32>()
      }
  }
  
  

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