#883 - Projection Area of 3D Shapes

Problem 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.

Solution

/**
 * @param {number[][]} grid
 * @return {number}
 */
var projectionArea = function(grid) {
  let topView = 0;
  let frontView = 0;
  let sideView = 0;

  for (let i = 0; i < grid.length; i++) {
    let maxRow = 0;
    let maxCol = 0;

    for (let j = 0; j < grid.length; j++) {
      if (grid[i][j] > 0) topView++;
      maxRow = Math.max(maxRow, grid[i][j]);
      maxCol = Math.max(maxCol, grid[j][i]);
    }

    frontView += maxRow;
    sideView += maxCol;
  }

  return topView + frontView + sideView;
};