LeetCode in Kotlin

812. Largest Triangle Area

Easy

Given an array of points on the X-Y plane points where points[i] = [xi, yi], return the area of the largest triangle that can be formed by any three different points. Answers within 10-5 of the actual answer will be accepted.

Example 1:

Input: points = [[0,0],[0,1],[1,0],[0,2],[2,0]]

Output: 2.00000

Explanation: The five points are shown in the above figure. The red triangle is the largest.

Example 2:

Input: points = [[1,0],[0,0],[0,1]]

Output: 0.50000

Constraints:

• 3 <= points.length <= 50
• -50 <= xi, yi <= 50
• All the given points are unique.

Solution

import kotlin.math.abs

class Solution {
    fun largestTriangleArea(points: Array<IntArray>): Double {
        val n = points.size
        var max = 0.0
        for (i in 0 until n) {
            for (j in i + 1 until n) {
                for (k in j + 1 until n) {
                    var area: Double
                    val a = points[i]
                    val b = points[j]
                    val c = points[k]
                    area = abs(area(a, b) + area(b, c) + area(c, a))
                    if (area > max) {
                        max = area
                    }
                }
            }
        }
        return max
    }

    private fun area(a: IntArray, b: IntArray): Double {
        val l = b[0] - a[0]
        val h = (a[1] + b[1] + 200) / 2.0
        return l * h
    }
}

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