Easy
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
public class Solution {
public int projectionArea(int[][] grid) {
int n = grid.length;
int m = grid[0].length;
int sum = n * m;
int count = 0;
for (int[] ints : grid) {
int max = Integer.MIN_VALUE;
for (int j = 0; j < m; j++) {
if (ints[j] == 0) {
count++;
}
if (max < ints[j]) {
max = ints[j];
}
}
sum += max;
}
for (int i = 0; i < n; i++) {
int max = Integer.MIN_VALUE;
for (int j = 0; j < m; j++) {
if (max < grid[j][i]) {
max = grid[j][i];
}
}
sum += max;
}
return sum - count;
}
}