Projection area of 3d shapes

Array Math Geometry Matrix

75.6% acceptance

Feb 22, 2026

638

1451

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

Rust

Time O(n²)

Space O(1)

LeetCode

solution.rs

/*
 * 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
 */

impl Solution {
  pub fn projection_area(grid: Vec<Vec<i32>>) -> i32 {
    let n = grid.len();
    let mut xy = 0i32; // top
    let mut xz = 0i32; // front
    let mut yz = 0i32; // side
    for i in 0..n {
      let mut max_row = 0i32;
      let mut max_col = 0i32;
      for j in 0..n {
        if grid[i][j] > 0 { xy += 1; }
        max_row = max_row.max(grid[i][j]);
        max_col = max_col.max(grid[j][i]);
      }
      xz += max_row;
      yz += max_col;
    }
    xy + xz + yz
  }
}