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Leetcode 812. Largest Triangle Area

Problem Statement

Given a list of points in the 2D plane, find the area of the largest triangle that can be formed by any three of the points.

You may assume that all the points lie within the range [-50, 50] and all coordinates are integers.

Your answer should be within 10^-6 of the true answer.

Clarifying Questions

  1. Input format: What is the structure of the input data?
    • The input is a list of points, where each point is represented as [x, y].
  2. Output format: What is the structure of the output?
    • The output should be a single floating-point number representing the maximum area, within 10^-6 of the true answer.
  3. Constraints: Are there constraints on the number of points?
    • The number of points (n) will be between 3 and 50 inclusive.
  4. Edge cases: Do we need to consider edge cases like all points being collinear?
    • Yes, we should ensure the solution handles cases where some or all points might be collinear.

Strategy

  1. Combinatorial Approach:
    • We’ll use a combinatorial approach to check every possible triplet of points and calculate the area of the triangle formed by these points.
  2. Area Calculation:
    • To calculate the area of a triangle given its vertices ((x1, y1)), ((x2, y2)), and ((x3, y3)), we can use the Shoelace formula: [ \text{Area} = \frac{1}{2} \left| x1 \cdot (y2 - y3) + x2 \cdot (y3 - y1) + x3 \cdot (y1 - y2) \right| ]
  3. Brute-force Optimization:
    • Given that the maximum number of points is 50, the number of combinations for triplets (( \binom{50}{3} )) is manageable for a brute-force solution.
  4. Floating-Point Precision:
    • Ensure the computation adheres to the precision requirements.

Code

Here is the implementation in Java:

public class Solution {
    public double largestTriangleArea(int[][] points) {
        int n = points.length;
        double maxArea = 0.0;

        for (int i = 0; i < n - 2; i++) {
            for (int j = i + 1; j < n - 1; j++) {
                for (int k = j + 1; k < n; k++) {
                    double area = calculateArea(points[i], points[j], points[k]);
                    maxArea = Math.max(maxArea, area);
                }
            }
        }
        return maxArea;
    }

    private double calculateArea(int[] p1, int[] p2, int[] p3) {
        // Using the Shoelace formula to calculate area of a triangle
        return 0.5 * Math.abs(p1[0] * (p2[1] - p3[1]) + p2[0] * (p3[1] - p1[1]) + p3[0] * (p1[1] - p2[1]));
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        int[][] points = \ use example from above
        System.out.println(sol.largestTriangleArea(points)); // Output should be 2.0
    }
}

Time Complexity

The time complexity for this solution is (O(n^3)), where (n) is the number of points. This is because we are considering all combinations of three points from the list, which is (\binom{n}{3} \approx \frac{n^3}{6}).

The space complexity is (O(1)) beyond the input storage, as we are using only a few extra variables to store computed areas and the maximum area value.

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