• 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

Solution: Geometry

Step 1: Build a largest pyramid that has less then n cubes, whose base area is d*(d+1) / 2
Step 2: Build a largest triangle with cubes left, whose base area is l, l*(l + 1) / 2 >= left

Time complexity: O(n^(1/3))
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int minimumBoxes(int n) {
    int d = 0;
    int l = 0;
    while (n - (d + 1) * (d + 2) / 2 > 0) {
      n -= (d + 1) * (d + 2) / 2;
      ++d;
    }
    while (n > 0) n -= ++l;
    return d * (d + 1) / 2 + l;
  }
};

The post 花花酱 LeetCode 1739. Building Boxes appeared first on Huahua's Tech Road.

If it is impossible to form any triangle of non-zero area, return 0.

Example 1:

Input: [2,1,2]
Output: 5

Example 2:

Input: [1,2,1]
Output: 0

Example 3:

Input: [3,2,3,4]
Output: 10

Example 4:

Input: [3,6,2,3]
Output: 8

Note:

1. 3 <= A.length <= 10000
2. 1 <= A[i] <= 10^6

Solution: Greedy

Answer must be 3 consecutive numbers in the sorted array
if A[i] >= A[i+1] + A[i+2], then A[i] >= A[i+j] + A[i+k], 1 < j < k
if A[i] < A[i+1] + A[i+2], then A[i] + A[i+1] + A[i+2] is the answer

class Solution {
public:
  int largestPerimeter(vector<int>& A) {
    sort(rbegin(A), rend(A));
    for (int i = 0; i < A.size() - 2; ++i)
      if (A[i] < A[i + 1] + A[i + 2])
        return A[i] + A[i + 1] + A[i + 2];
    return 0;
  }
};

The post 花花酱 LeetCode 976. Largest Perimeter Triangle appeared first on Huahua's Tech Road.

题目大意：给你一些点，求用这些点能够成的最大的三角形面积是多少？

https://leetcode.com/problems/largest-triangle-area/description/

You have a list of points in the plane. Return the area of the largest triangle that can be formed by any 3 of the points.

Example:
Input: points = [[0,0],[0,1],[1,0],[0,2],[2,0]]
Output: 2
Explanation: 
The five points are show in the figure below. The red triangle is the largest.

Notes:

• 3 <= points.length <= 50.
• No points will be duplicated.
•  -50 <= points[i][j] <= 50.
• Answers within 10^-6 of the true value will be accepted as correct.

Solution: Brute Force

Time complexity: O(n^3)

Space complexity: O(1)

// Author: Huahua
// Running time: 11 ms
class Solution {
public:
  double largestTriangleArea(vector<vector<int>>& points) {
    const int n = points.size();
    double best = 0.0;
    for (int i = 0; i < n; ++i)
      for (int j = i + 1; j < n; ++j)
        for (int k = j + 1; k < n; ++k) {          
          const double a = dist(points[i], points[j]);
          const double b = dist(points[i], points[k]);
          const double c = dist(points[j], points[k]);
          const double s = (a + b + c) / 2;
          const double area = sqrt(s * (s - a) * (s - b) * (s - c));          
          best = max(best, area);
        }
    return best;
  }
private:
  static inline double dist(const vector<int>& p1, const vector<int>& p2) {
    return sqrt((p1[0] - p2[0]) * (p1[0] - p2[0]) + (p1[1] - p2[1]) * (p1[1] - p2[1]));
  }
};

The post 花花酱 LeetCode 812. Largest Triangle Area appeared first on Huahua's Tech Road.