Building Boxes

The problem can be solved by visualizing how the boxes can be arranged in layers. Each layer forms a triangular base.

A triangular number can be represented as ( T_k = \frac{k \cdot (k + 1)}{2} ), and it describes the total number of boxes in ( k ) layers.

Let’s break down the problem into steps:

  1. Find the Complete Layers: Determine the maximum number of complete layers that can be formed. For this, you need to find the maximum ( k ) such that ( \frac{k \cdot (k + 1) \cdot (k + 2)}{6} \leq n ).
  2. Handle Remaining Boxes: If there are any remaining boxes, place them on top of the previous layers. You’ll need to calculate how many more boxes can be placed on the floor in the next layer without forming a complete layer.
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class Solution:
    def minimumBoxes(self, n: int) -> int:
        # Step 1: Find the complete layers
        k = 0
        total_boxes = 0
        while total_boxes + (k + 1) * (k + 2) // 2 <= n:
            k += 1
            total_boxes += k * (k + 1) // 2

        # Step 2: Handle remaining boxes
        remaining_boxes = n - total_boxes
        floor_boxes = k * (k + 1) // 2  # Boxes on the floor in complete layers
        i = 1
        while remaining_boxes > 0:
            remaining_boxes -= i
            floor_boxes += 1
            i += 1

        return floor_boxes

This code finds the minimum number of boxes touching the floor for a given ( n ). It efficiently calculates the number of complete layers and then carefully places any remaining boxes on top.