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:
![](https://spcdn.pages.dev/leetcode/problems/0883.Projection%20Area%20of%203D%20Shapes/images/shadow.png)
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#
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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
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| 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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| 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;
}
};
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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
}
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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>()
}
}
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