892. Surface Area of 3D Shapes LeetCode Solution

Last updated: February 20, 2025

In this guide, you will get 892. Surface Area of 3D Shapes LeetCode Solution with the best time and space complexity. The solution to Surface Area of D Shapes problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Surface Area of D Shapes solution in C++
Surface Area of D Shapes solution in Java
Surface Area of D Shapes solution in Python
Additional Resources

892. Surface Area of 3D Shapes LeetCode Solution image

Problem Statement of Surface Area of D Shapes

You are given an n x n grid where you have placed some 1 x 1 x 1 cubes. Each value v = grid[i][j] represents a tower of v cubes placed on top of cell (i, j).
After placing these cubes, you have decided to glue any directly adjacent cubes to each other, forming several irregular 3D shapes.
Return the total surface area of the resulting shapes.
Note: The bottom face of each shape counts toward its surface area.

Example 1:

Input: grid = [[1,2],[3,4]]
Output: 34

Example 2:

Input: grid = [[1,1,1],[1,0,1],[1,1,1]]
Output: 32

Example 3:

Input: grid = [[2,2,2],[2,1,2],[2,2,2]]
Output: 46

Constraints:

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

Complexity Analysis

Time Complexity:
Space Complexity:

892. Surface Area of 3D Shapes LeetCode Solution in C++

class Solution {
 public:
  int surfaceArea(vector<vector<int>>& grid) {
    int ans = 0;

    for (int i = 0; i < grid.size(); ++i)
      for (int j = 0; j < grid.size(); ++j) {
        if (grid[i][j])
          ans += grid[i][j] * 4 + 2;
        if (i > 0)
          ans -= min(grid[i][j], grid[i - 1][j]) * 2;
        if (j > 0)
          ans -= min(grid[i][j], grid[i][j - 1]) * 2;
      }

    return ans;
  }
};
/* code provided by PROGIEZ */

892. Surface Area of 3D Shapes LeetCode Solution in Java

class Solution {
  public int surfaceArea(int[][] grid) {
    int ans = 0;

    for (int i = 0; i < grid.length; ++i)
      for (int j = 0; j < grid.length; ++j) {
        if (grid[i][j] > 0)
          ans += grid[i][j] * 4 + 2;
        if (i > 0)
          ans -= Math.min(grid[i][j], grid[i - 1][j]) * 2;
        if (j > 0)
          ans -= Math.min(grid[i][j], grid[i][j - 1]) * 2;
      }

    return ans;
  }
}
// code provided by PROGIEZ

892. Surface Area of 3D Shapes LeetCode Solution in Python

class Solution:
  def surfaceArea(self, grid: list[list[int]]) -> int:
    ans = 0

    for i in range(len(grid)):
      for j in range(len(grid)):
        if grid[i][j]:
          ans += grid[i][j] * 4 + 2
        if i > 0:
          ans -= min(grid[i][j], grid[i - 1][j]) * 2
        if j > 0:
          ans -= min(grid[i][j], grid[i][j - 1]) * 2

    return ans
 # code by PROGIEZ

Additional Resources

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