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Building Boxes

Number: 1861

Difficulty: Hard

Paid? No

Companies: N/A

Problem Description

We are given n unit cubes and a cubic room (all sides = n). The cubes must be “stacked” following a rule: a box placed on top of another must have its underlying box “supported” on all four vertical sides – meaning every side of the lower box touches either an adjacent box or a wall. The goal is to arrange all n boxes (by placing them anywhere on the floor and then stacking on top when possible) so that the number of boxes on the floor is minimized. You must return the minimum possible number of boxes touching the floor.

Key Insights

  • The optimal stacking uses a “staircase” arrangement in the corner of the room, so the walls provide support.
  • Instead of placing boxes in a full rectangle on the floor, an optimal floor placement is a right‑angled triangle.
  • If the floor placement is arranged in m rows (first row has m boxes, next row m‑1, …, last row has 1 box), then the number of floor boxes is T = m(m+1)/2.
  • With the rules, it is possible to stack additional layers above the floor. In particular, if the floor is arranged as described, the maximum number of boxes that can be stacked in total is   total(m) = 1/6 · m · (m+1) · (m+2)
  • Our goal is to pick the smallest integer m (which determines the floor count T) so that total(m) ≥ n.
  • Once m is found, the answer is T = m(m+1)/2 (the number of floor boxes).

Space and Time Complexity

Time Complexity: O(log n) – we use binary search over m in a range that will be roughly O(n^(1/3)), which is very small relative to n. Space Complexity: O(1) – only a few variables are needed.

Solution

The idea is to “guess” the number m (the number of rows in the triangular floor placement) and check if using that many rows we can stack at least n boxes. To that end, we use the derived formula:   total(m) = m * (m+1) * (m+2) / 6 We perform a binary search for the smallest m such that total(m) ≥ n. Once found, the minimum number of boxes touching the floor is simply floorBoxes = m * (m+1) / 2. Key details:

  • We start m from 1.
  • The upper-bound for m can be determined loosely; note that total(m) grows as ~m³/6 so m is roughly (6*n)^(1/3).
  • Then we return the computed floorBoxes.

Code Solutions

Below are implementations in Python, JavaScript, C++, and Java.

# Python implementation
def minimumBoxes(n: int) -> int:
    # Helper function to compute the maximum boxes that
    # can be stacked given m rows in a triangular floor arrangement.
    def total_boxes(m: int) -> int:
        return m * (m + 1) * (m + 2) // 6

    # Binary search to find smallest m with total_boxes(m) >= n.
    lo, hi = 1, int((6 * n) ** (1/3)) + 2  # a safe upper bound
    while lo < hi:
        mid = (lo + hi) // 2
        if total_boxes(mid) < n:
            lo = mid + 1
        else:
            hi = mid
    m = lo
    # Number of floor boxes for a triangular arrangement is m(m+1)/2.
    return m * (m + 1) // 2

# Example usage:
print(minimumBoxes(3))  # Expected output: 3
print(minimumBoxes(4))  # Expected output: 3
print(minimumBoxes(10)) # Expected output: 6
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