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Building Boxes

Difficulty: Hard

Problem Description

You have a cubic storeroom where the width, length, and height of the room are all equal to n units. You are asked to place n boxes in this room where each box is a cube of unit side length. There are however some rules to placing the boxes:

  • You can place the boxes anywhere on the floor.
  • If box x is placed on top of the box y, then each side of the four vertical sides of the box y must either be adjacent to another box or to a wall.

Given an integer n, return the minimum possible number of boxes touching the floor.

Key Insights

  • Each box must be placed such that its vertical sides are supported by adjacent boxes or walls.
  • The optimal arrangement involves minimizing the number of boxes touching the floor while maximizing the use of vertical space.
  • A corner placement allows for efficient stacking of boxes.

Space and Time Complexity

Time Complexity: O(1)
Space Complexity: O(1)

Solution

To solve the problem, we can use a mathematical approach rather than simulating the placement of boxes. The key insight is to recognize that the number of boxes touching the floor can be minimized by maximizing the height of stacks in the corners of the room.

  1. For any given n, determine how many full layers of boxes can be placed.
  2. Calculate the total number of boxes that can be placed on the floor based on the largest square that can fit in the corner.
  3. The minimum number of boxes touching the floor will be bounded by the largest integer k such that k^2 + k(k-1) ≤ n, where k represents the number of layers.

This leads us to a mathematical formula involving the use of binary search to find the appropriate k for large values of n efficiently.

Code Solutions

def minBoxes(n: int) -> int:
    # Binary search to find the maximum k
    left, right = 1, n
    while left < right:
        mid = (left + right) // 2
        # Calculate the number of boxes with k layers
        if mid * (mid + 1) * (mid + 2) // 6 <= n:
            left = mid + 1
        else:
            right = mid
    return left - 1

# Example usage:
print(minBoxes(3))  # Output: 3
print(minBoxes(4))  # Output: 3
print(minBoxes(10))  # Output: 6
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