Find the total area covered by two rectilinear rectangles in a 2D plane.

Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.

Example:

Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9, H = 2
Output: 45

Note:

Assume that the total area is never beyond the maximum possible value of int.

Solution:

area1 + area2 – overlapped area

Time complexity: O(1)

Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
    int area1 = (C - A) * (D - B);
    int area2 = (G - E) * (H - F);
    int x1 = max(A, E);
    int x2 = max(x1, min(C, G));
    int y1 = max(B, F);
    int y2 = max(y1, min(D, H));
    return area1 + area2 - (x2 - x1) * (y2 - y1);
  }
};

// Author: Huahua
class Solution {
  public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
    int area1 = (C - A) * (D - B);
    int area2 = (G - E) * (H - F);
    int x1 = Math.max(A, E);
    int x2 = Math.max(x1, Math.min(C, G));
    int y1 = Math.max(B, F);
    int y2 = Math.max(y1, Math.min(D, H));
    return area1 + area2 - (x2 - x1) * (y2 - y1);
  }
}

# Author: Huahua
class Solution:
  def computeArea(self, A, B, C, D, E, F, G, H):
    area1 = (C - A) * (D - B);
    area2 = (G - E) * (H - F);
    x1 = max(A, E);
    x2 = max(x1, min(C, G));
    y1 = max(B, F);
    y2 = max(y1, min(D, H));
    return area1 + area2 - (x2 - x1) * (y2 - y1)

The post 花花酱 LeetCode 223. Rectangle Area appeared first on Huahua's Tech Road.

On a N * N grid, we place some 1 * 1 * 1 cubes.

Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell (i, j).

Return the total surface area of the resulting shapes.

Example 1:

Input: [[2]]
Output: 10

Example 2:

Input: [[1,2],[3,4]]
Output: 34

Example 3:

Input: [[1,0],[0,2]]
Output: 16

Example 4:

Input: [[1,1,1],[1,0,1],[1,1,1]]
Output: 32

Example 5:

Input: [[2,2,2],[2,1,2],[2,2,2]]
Output: 46

Note:

• 1 <= N <= 50
• 0 <= grid[i][j] <= 50

Solution: Geometry

3D version of 花花酱 LeetCode 463. Island Perimeter

Ans = total faces – hidden faces.

each pile with height h has the surface area of 2 (top/bottom) + 4*h (sides)

if a cube has a neighbour, reduce the total surface area by 1

For each pile, we check 4 neighbours, the number of total hidden faces of this pile is sum(min(h, neighbour’s h)).

Time complexity: O(mn)

// Author: Huahua
// Running time: 4 ms
class Solution {
public:
  int surfaceArea(vector<vector<int>>& grid) {
    static const vector<int> dirs{0, -1, 0, 1, 0};
    int m = grid.size();
    int n = grid[0].size();
    int ans = 0;
    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j) {
        int h = grid[i][j];
        if (h == 0) continue;
        ans += 2 + 4 * h;        
        for (int k = 0; k < 4; k++) {
          int tx = j + dirs[k];
          int ty = i + dirs[k + 1];
          if (tx < 0 || tx >= n || ty < 0 || ty >= m) continue;
          int th = grid[ty][tx];          
          ans -= (th <= 0 ? 0 : min(h, th));
        }
      }
    return ans;
  }
};

// Author: Huahua
// Running time: 4 ms
class Solution {
public:
  int surfaceArea(vector<vector<int>>& grid) {    
    int m = grid.size();
    int n = grid[0].size();
    int ans = 0;
    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j) {
        int h = grid[i][j];
        if (h == 0) continue;
        ans += 2 + 4 * h;
        if (i) ans -= 2 * min(h, grid[i - 1][j]);
        if (j) ans -= 2 * min(h, grid[i][j - 1]);        
      }
    return ans;
  }
};

The post 花花酱 LeetCode 892. Surface Area of 3D Shapes 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.

Problem:

Given n non-negative integers a₁, a₂, …, a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

Examples:

input: [1 3 2 4]
output: 6
explanation: use 3, 4, we have the following, which contains (4th-2nd) * min(3, 4) = 2 * 3 = 6 unit of water.

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|~~|

Idea:

Two pointers

Time complexity: O(n)

Space complexity: O(1)

Solution:

C++ two pointers

// Author: Huahua
// Runtime: 19 ms
class Solution {
public:
    int maxArea(const vector<int>& height) {
        int ans = 0;
        int l = 0;
        int r = height.size() - 1;
        while (l < r) {
            int h = min(height[l], height[r]);
            ans = max(ans, h * (r - l));
            if (height[l] < height[r]) 
                ++l;
            else
                --r;
        }
        return ans;
    }
};

Java

// Author: Huahua
// Runtime: 13 ms
class Solution {
    public int maxArea(int[] height) {
        int l = 0;
        int r = height.length - 1;
        int ans = 0;
        while (l < r) {
            int h = Math.min(height[l], height[r]);
            ans = Math.max(ans, h * (r - l));
            if (height[l] < height[r])
                ++l;
            else
                --r;
        }
        return ans;
    }
}

The post 花花酱 LeetCode 11. Container With Most Water appeared first on Huahua's Tech Road.