Largest Triangle Area

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- Tag: math, #triangle

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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Related Reading: Shoelace formula in wikipedia

## https://code.dennyzhang.com/largest-triangle-area ## Basic Ideas: ## Check whether 3 points can form a triangle ## any 2 edges should be bigger than the 3rd one ## Can't be in one line: k = (y2-y1)/(x2-x1) ## Get area of a triangle from 3 edges ## https://www.mathsisfun.com/geometry/herons-formula.html ## s = (a+b+c)/2, area = sqrt(s*(s-a)*(s-b)*(s-c)) ## Complexity: Time O(n^3), Space O(1) class Solution: def largestTriangleArea(self, points): """ :type points: List[List[int]] :rtype: float """ from math import sqrt def getArea(point1, point2, point3): # get 3 edges a = sqrt(pow((point1[0]-point2[0]), 2) + pow((point1[1]-point2[1]), 2)) b = sqrt(pow((point1[0]-point3[0]), 2) + pow((point1[1]-point3[1]), 2)) c = sqrt(pow((point2[0]-point3[0]), 2) + pow((point2[1]-point3[1]), 2)) # same point if 0 in [a, b, c]: return 0 # valid edges if a+b<=c or a+c<=b or b+c<=a: return 0 # check whether 3 points in the same line if point1[0]-point2[0] == 0: k1 = None else: k1 = (point1[1]-point2[1])/(point1[0]-point2[0]) if point1[0]-point3[0] == 0: k2 = None else: k2 = (point1[1]-point3[1])/(point1[0]-point3[0]) if point2[0]-point3[0] == 0: k3 = None else: k3 = (point2[1]-point3[1])/(point2[0]-point3[0]) if len(set([k1, k2, k3])) != 3: return 0 # get area s = (a+b+c)/2 area = sqrt(s*(s-a)*(s-b)*(s-c)) return area res = 0 for i in range(len(points)-2): for j in range(i+1, len(points)-1): for k in range(j+1, len(points)): res = max(res, getArea(points[i], points[j], points[k])) return res