# 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](https://www.chatmagic.app), 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
	```python
	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
	```