812. Largest Triangle Area

A = (1/2) [x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)]

T: O(n^3)

```java
class Solution {
    public double largestTriangleArea(int[][] points) {
        double result = 0;
        for (int i = 0; i < points.length; i++) {
            for (int j = i+1; j < points.length; j++) {
                for (int k = j+1; k < points.length; k++) {
                    result = Math.max(result, helper(points[i], points[j], points[k]));
                }
            }
        }
        return result;
    }
    private double helper(int num1[], int num2[], int num3[]) {
        double x1 = (double) num1[0];
        double y1 = (double) num1[1];

        double x2 = (double) num2[0];
        double y2 = (double) num2[1];
        
        double x3 = (double) num3[0];
        double y3 = (double) num3[1];

        return Math.abs((x1*(y2-y3) + x2*(y3 - y1) + x3*(y1 - y2)))/2.0;
    }
}

/**
who knows it, fuck

using Shoelace's Formula (also known as Gauss's area formula and surveyor's formula):
Area = 0.5 * |x1*(y2-y3) + x2*(y3-y1) + x3(y1-y2)|
 */
```

public double largestTriangleArea(int[][] p) {
        double res = 0;
        for (int[] i: p)
            for (int[] j: p)
                for (int[] k: p)
            res = Math.max(res, 0.5 * Math.abs(
            i[0] * j[1] + j[0] * k[1] + k[0] * i[1]
            - j[0] * i[1] - k[0] * j[1] - i[0] * k[1])
            );
        return res;
    }