892. Surface Area of 3D Shapes
Explanation
To solve this problem, we can iterate through each cell in the grid and calculate the contribution of each cell to the total surface area. The surface area of a single cube is 6, and we need to consider the surface area of each cell as well as the shared faces with adjacent cells. We will calculate the total surface area by subtracting the areas of the shared faces.
- Iterate through each cell in the grid.
- For each cell, calculate the surface area contributed by that cell (4 sides * height) and add it to the total surface area.
- For each cell, check its adjacent cells to subtract the shared faces. If the adjacent cell has a higher value, subtract the overlapping faces. If the adjacent cell has a lower value, do not subtract anything (as the face is already accounted for in the adjacent cell's calculation).
- Return the total surface area.
The time complexity of this solution is O(n^2) as we are iterating through each cell in the grid. The space complexity is O(1) as we are using only a constant amount of extra space.
class Solution {
public int surfaceArea(int[][] grid) {
int n = grid.length;
int surfaceArea = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] > 0) {
surfaceArea += 4 * grid[i][j] + 2;
if (i > 0) {
surfaceArea -= 2 * Math.min(grid[i][j], grid[i - 1][j]);
}
if (j > 0) {
surfaceArea -= 2 * Math.min(grid[i][j], grid[i][j - 1]);
}
}
}
}
return surfaceArea;
}
}Code Editor (Testing phase)
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