# Projection Area Of 3D Shapes Problem

## Description

LeetCode Problem 883.

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:

``````1
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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:

``````1
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Input: grid = [[2]]
Output: 5
``````

Example 3:

``````1
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Input: grid = [[1,0],[0,2]]
Output: 8
``````

Example 4:

``````1
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Input: grid = [[1,1,1],[1,0,1],[1,1,1]]
Output: 14
``````

Example 5:

``````1
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Input: grid = [[2,2,2],[2,1,2],[2,2,2]]
Output: 21
``````

Constraints:

• n == grid.length
• n == grid[i].length
• 1 <= n<= 50
• 0 <= grid[i][j] <= 50

## Sample C++ Code

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class Solution {
public:
int projectionArea(vector<vector<int>>& grid) {
int area = 0;
int row = grid.size();
int col = grid[0].size();
int maxvalrow = INT_MIN;
int maxvalcol = INT_MIN;
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
maxvalcol = max(maxvalcol, grid[i][j]);
maxvalrow = max(maxvalrow, grid[j][i]);
if (grid[i][j])
area++;
}
area += maxvalcol;
area += maxvalrow;
maxvalrow = INT_MIN;
maxvalcol = INT_MIN;
}
return area;
}
};
``````