comments | difficulty | edit_url | tags | ||||
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Easy |
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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
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 grid
. The space complexity is
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
}
}