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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.
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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; } }
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// 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; } };
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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
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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 }
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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; }
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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>() } }