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Projection Area of 3D Shapes

Updated on 20 June, 2025
Projection Area of 3D Shapes header image

Problem Statement

Imagine a square n x n grid where every cell can hold a stack of 1 x 1 x 1 cubes. The number of cubes in each cell (i, j) is represented by the integer value v = grid[i][j]. The task is to visualize and calculate the total projected area of these stacks on three different planes: the top (xy plane), the front (yz plane), and the side (zx plane).

The concept of projection used here can be likened to observing the shadow cast by these cubes when light is shone from above, the front, and the side, corresponding to observing the grid from three different orthogonal views. Each view will project a different shadow based on the height and configuration of cubes in the grid. Your goal is to compute the collective area from all three projections.

Examples

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

Approach and Intuition

To solve this problem effectively, understanding what exactly each projection entails and how to calculate their areas becomes vital:

  1. Top Projection (XY plane):

    • The top projection will simply be the count of all cells that have at least one cube. In essence, every non-zero cell in the grid adds 1 to the projection's area on this plane.

  2. Front Projection (YZ plane):

    • This view is observed by looking at the grid from one side (say, from the left or right). Therefore, only the tallest stack (the maximum value) in each column determines the shadow for that particular column. Summing up the maximum heights from each column gives the total projection area on this plane.

  3. Side Projection (ZX plane):

    • Similarly, for the side projection (observed from top or bottom of the grid), the tallest stack in each row casts the shadow. Here, you need to compute the maximum stack in each row and sum these to get the total area for this projection.

By correctly calculating the area for each projection and summing these areas, you yield the total projected area of these 3D stacks of cubes as viewed from the top, front, and side. This approach leverages basic matrix traversal and is efficient given the constraints.

Solutions

  • C++
  • Java
cpp 
class Solution {
public:
    int calculateProjectionArea(vector<vector<int>>& matrix) {
        int size = matrix.size();
        int totalArea = 0;
    
        for (int i = 0; i < size; ++i) {
            int rowMax = 0;
            int colMax = 0;
            for (int j = 0; j < size; ++j) {
                if (matrix[i][j] > 0) totalArea++;
                rowMax = max(rowMax, matrix[i][j]);
                colMax = max(colMax, matrix[j][i]);
            }
            totalArea += rowMax + colMax;
        }
    
        return totalArea;
    }
};

This summary provides a solution in C++ for calculating the projection area of 3D shapes represented as matrices. The chief intent of the solution is to determine the sum of the top view, front view, and side view areas derived from the grid values of the 3D shape.

  • First, intialize size to capture the dimension of the matrix, and totalArea to store the result.
  • Loop through each row index i in the matrix. Within this loop:
    • Initialize rowMax and colMax to zero. These will store the maximum values for each row and column, respectively, which are essential to determine the front view and side view.
    • Begin another loop over each column index j:
      • Increase totalArea for each non-zero cell encountered. This increment accounts for the top view projection, where every non-zero value contributes to the area.
      • Update rowMax with the maximum value in the current row by comparing it with matrix[i][j].
      • Similarly, update colMax with the maximum value in the current column by comparing it with matrix[j][i].
    • After processing each row and column, add the maximum values of the current row and column to totalArea.
  • Finally, return totalArea as the result. This value represents the combined projection area of the top, front, and side views of the 3D shape.

Ensure that you include the necessary header files and use appropriate data types for variable declarations to implement this efficiently in C++.

java 
class Solution {
    public int calculateProjectionArea(int[][] matrix) {
        int size = matrix.length;
        int total = 0;
    
        for (int i = 0; i < size; i++) {
            int maxRow = 0; 
            int maxCol = 0; 
            for (int j = 0; j < size; j++) {
                if (matrix[i][j] > 0) total++;  
                maxRow = Math.max(maxRow, matrix[i][j]);
                maxCol = Math.max(maxCol, matrix[j][i]);
            }
            total += maxRow + maxCol;
        }
    
        return total;
    }
}

This solution involves calculating the projection area of a 3D shape on the x-y, y-z, and x-z planes using given 3D grid values represented as a 2D matrix.

Here's how the solution works:

  • Initialize size to the length of the input matrix, which represents the dimensions of the grid.
  • Initialize total to 0, which will hold the sum of the projection areas onto the three planes.
  • Iterate through each cell of the matrix using two nested loops. The outer loop iterates over rows and the inner loop iterates over columns.
  • Initialize maxRow and maxCol to zero for each iteration of the outer loop. These variables will keep track of the maximum values in the current row and column, respectively.
  • In the inner loop:
    • Increase the total by one for each non-zero cell to account for the top view projection onto the x-y plane.
    • Update maxRow with the maximum value found in the current row up to the current column.
    • Update maxCol with the maximum value found in the current column up to the current row.
  • After exiting the inner loop, add the maximum values from maxRow and maxCol to total, representing the projections on the y-z and x-z planes for that particular row and column.
  • Return total, which by now accumulates all three projection areas.

This approach efficiently loops through the matrix, and at each step, computations focus on relevant maximums and counts needed for the final area calculation. Using minimal space and only double traversal of the matrix, the solution ensures optimal performance suitable for varying sizes of input matrices.

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