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Formatted question description: https://leetcode.ca/all/883.html

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.

  • class Solution {
        public int projectionArea(int[][] grid) {
            int areaTop = 0, areaFront = 0, areaSide = 0;
            int rows = grid.length, columns = grid[0].length;
            int[] maxEachRow = new int[rows];
            int[] maxEachColumn = new int[columns];
            for (int i = 0; i < rows; i++) {
                for (int j = 0; j < columns; j++) {
                    int cell = grid[i][j];
                    if (cell != 0)
                        areaTop++;
                    maxEachRow[i] = Math.max(maxEachRow[i], cell);
                    maxEachColumn[j] = Math.max(maxEachColumn[j], cell);
                }
            }
            for (int num : maxEachRow)
                areaSide += num;
            for (int num : maxEachColumn)
                areaFront += num;
            return areaTop + areaFront + areaSide;
        }
    }
    
    ############
    
    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;
        }
    }
    
  • // OJ: https://leetcode.com/problems/projection-area-of-3d-shapes/
    // 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;
        }
    };
    
  • 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
    
    ############
    3
    class Solution(object):
        def projectionArea(self, grid):
            """
            :type grid: List[List[int]]
            :rtype: int
            """
            top, front, side = 0, 0, 0
            n = len(grid)
            for i in range(n):
                x, y = 0, 0
                for j in range(n):
                    if grid[i][j] != 0:
                        top += 1
                    x = max(x, grid[i][j])
                    y = max(y, grid[j][i])
                front += x
                side += y
            return top + front + side
    
  • 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
    }
    
    func max(a, b int) int {
    	if a > b {
    		return a
    	}
    	return b
    }
    
  • 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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