README_EN.md

1739. Building Boxes

中文文档

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.

 

Example 1:

Input: n = 3
Output: 3
Explanation: The figure above is for the placement of the three boxes.
These boxes are placed in the corner of the room, where the corner is on the left side.

Example 2:

Input: n = 4
Output: 3
Explanation: The figure above is for the placement of the four boxes.
These boxes are placed in the corner of the room, where the corner is on the left side.

Example 3:

Input: n = 10
Output: 6
Explanation: The figure above is for the placement of the ten boxes.
These boxes are placed in the corner of the room, where the corner is on the back side.

 

Constraints:

• 1 <= n <= 109

Solutions

Python3

class Solution:
    def minimumBoxes(self, n: int) -> int:
        s, k = 0, 1
        while s + k * (k + 1) // 2 <= n:
            s += k * (k + 1) // 2
            k += 1
        k -= 1
        ans = k * (k + 1) // 2
        k = 1
        while s < n:
            ans += 1
            s += k
            k += 1
        return ans

Java

class Solution {
    public int minimumBoxes(int n) {
        int s = 0, k = 1;
        while (s + k * (k + 1) / 2 <= n) {
            s += k * (k + 1) / 2;
            ++k;
        }
        --k;
        int ans = k * (k + 1) / 2;
        k = 1;
        while (s < n) {
            ++ans;
            s += k;
            ++k;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int minimumBoxes(int n) {
        int s = 0, k = 1;
        while (s + k * (k + 1) / 2 <= n) {
            s += k * (k + 1) / 2;
            ++k;
        }
        --k;
        int ans = k * (k + 1) / 2;
        k = 1;
        while (s < n) {
            ++ans;
            s += k;
            ++k;
        }
        return ans;
    }
};

Go

func minimumBoxes(n int) int {
	s, k := 0, 1
	for s+k*(k+1)/2 <= n {
		s += k * (k + 1) / 2
		k++
	}
	k--
	ans := k * (k + 1) / 2
	k = 1
	for s < n {
		ans++
		s += k
		k++
	}
	return ans
}

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