883. Projection Area of 3D Shapes

Problem

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

Solution (Java)

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;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).