Solving Leetcode Interviews in Seconds with AI: Projection Area of 3D Shapes
Introduction
In this blog post, we will explore how to solve the LeetCode problem "883" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.
Problem Statement
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
Explanation
Here's the breakdown of the solution:
- Top Projection (XY-plane): The area of the top projection is simply the number of cells in the grid that have at least one cube (i.e.,
grid[i][j] > 0). - Front Projection (XZ-plane): For each row, the height of the projection is the maximum value in that row. Summing these maximum heights gives the front projection area.
Side Projection (YZ-plane): Similarly, for each column, the height of the projection is the maximum value in that column. Summing these maximum heights gives the side projection area.
Time & Space Complexity: O(n^2) time, O(1) space.
Code
def projectionArea(grid):
"""
Calculates the total area of the three projections of a 3D shape.
Args:
grid: A list of lists of integers representing the height of cubes at each cell.
Returns:
The total area of the three projections.
"""
n = len(grid)
top_area = 0
front_area = 0
side_area = 0
for i in range(n):
max_row = 0
max_col = 0
for j in range(n):
if grid[i][j] > 0:
top_area += 1
max_row = max(max_row, grid[i][j])
max_col = max(max_col, grid[j][i])
front_area += max_row
side_area += max_col
return top_area + front_area + side_area