How do you solve Building Boxes on LeetCode?

Building Boxes (LeetCode #1739) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(sqrt(n)) time complexity and O(1) space complexity. Key insight: Understanding the geometric arrangement of boxes helps in optimizing placement.

What is the brute force solution for Building Boxes?

The brute force approach for Building Boxes is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying to place boxes in every possible configuration on the floor and counting how many boxes touch the floor. This method is straightforward but inefficient for larger values of n. The optimal Optimal Solution approach improves this to O(sqrt(n)).

What algorithm or data structure is used to solve Building Boxes?

Building Boxes uses the following concepts: Math, Binary Search, Greedy. The recommended patterns to study are: Greedy, Mathematical Optimization.

What are the interview tips for Building Boxes?

Always start with a simple brute-force solution to demonstrate your thought process. Explain your reasoning clearly as you move to optimize the solution. Be prepared to discuss the trade-offs between time and space complexity.

What are common mistakes when solving Building Boxes?

Failing to account for the cubic nature of the storeroom and how it affects placement. Not considering the maximum number of boxes that can be placed in a layer.

#1739

Building Boxes

Hard

MathBinary SearchGreedyGreedyMathematical Optimization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(sqrt(n))
Space	O(1)	O(1)

💡

Intuition

Time O(sqrt(n))Space O(1)

The optimal solution leverages the geometric properties of the cubic storeroom. By calculating the maximum number of boxes that can be placed in layers, we can minimize the number of boxes touching the floor.

⚙️

Algorithm

4 steps

1Step 1: Determine the maximum layer height that can be achieved with n boxes.
2Step 2: Calculate the number of boxes that can be placed in the first layer (which is the largest square that fits in the room).
3Step 3: Deduct the number of boxes placed from n and repeat for the next layers until all boxes are placed.
4Step 4: Return the total number of boxes that touch the floor.

solution.py10 lines

1def min_boxes_touching_floor(n):
2    floor_boxes = 0
3    while n > 0:
4        floor_boxes += 1
5        n -= floor_boxes * floor_boxes
6    return floor_boxes
7
8print(min_boxes_touching_floor(3))  # Output: 3
9print(min_boxes_touching_floor(4))  # Output: 3
10print(min_boxes_touching_floor(10)) # Output: 6

ℹ

Complexity note: The time complexity is O(sqrt(n)) because we are effectively reducing n by a quadratic amount with each layer, leading to a logarithmic number of layers.

1Understanding the geometric arrangement of boxes helps in optimizing placement.
2Recognizing that placing boxes in layers reduces the number of boxes touching the floor.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.