1166EasyArrayMathGeometryPolygons

The Largest Triangle Between Pegs

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A board holds pegs at whole-number spots given by pegs, where pegs[i] = [x, y], and no two pegs share a spot.

Return the largest area of a triangle whose three corners are pegs. Three pegs standing in a straight line make a triangle of area zero.

Examples

Example 1

Input: pegs = [[0, 0], [3, 0], [0, 4]]
Output: 6.0

The three pegs make a right-angled triangle with sides of 3 and 4 along the axes, so the area is half of twelve.

Example 2

Input: pegs = [[0, 0], [1, 1], [2, 2]]
Output: 0.0

All three pegs stand on one line, so the only triangle available has no area.

Example 3

Input: pegs = [[-1, -1], [1, -1], [0, 1]]
Output: 2.0

The base runs 2 across and the far peg stands 2 above it, so the area is half of four.

Constraints

3 <= pegs.length <= 50
-50 <= pegs[i][0] <= 50
-50 <= pegs[i][1] <= 50
no two pegs share a spot

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def largest_triangle_area(pegs: list[list[int]]) -> float:

Java

public double largestTriangleArea(int[][] pegs)

The intended approach

Time O(n^3) over the triples of pegs.Space O(1).

With at most fifty pegs there are fewer than twenty thousand triples, so every triple can be measured outright. What matters is measuring one cheaply and exactly. The shoelace reading does it: for corners `a`, `b` and `c`, the quantity `ax(by - cy) + bx(cy - ay) + cx(ay - by)` is twice the signed area. It comes out of whole-number arithmetic alone, so nothing is lost to rounding, and the only care needed is to drop its sign, since going round the corners the other way flips it. Three pegs in a line give nothing at all, which is the right answer for them. Each triple is looked at once with the corners in increasing order, so no triangle is measured three times over, and halving happens at the end of each measurement.

Pythonentry point: largest_triangle_area

def largest_triangle_area(pegs):
    n = len(pegs)
    best = 0.0
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                ax = pegs[i][0]
                ay = pegs[i][1]
                bx = pegs[j][0]
                by = pegs[j][1]
                cx = pegs[k][0]
                cy = pegs[k][1]
                twice = ax * (by - cy) + bx * (cy - ay) + cx * (ay - by)
                if twice < 0:
                    twice = -twice
                area = twice / 2.0
                if area > best:
                    best = area
    return best

The obvious version

Take every ordered triple of different pegs, build the two edge vectors leading away from the first corner, and take the size of their cross product as twice the area. Costs the peg count cubed with every triangle measured six times over.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_largest_triangle_area(pegs):
    n = len(pegs)
    best = 0.0
    for i in range(n):
        for j in range(n):
            for k in range(n):
                if i == j or j == k or i == k:
                    continue
                ux = pegs[j][0] - pegs[i][0]
                uy = pegs[j][1] - pegs[i][1]
                vx = pegs[k][0] - pegs[i][0]
                vy = pegs[k][1] - pegs[i][1]
                cross = ux * vy - uy * vx
                if cross < 0:
                    cross = -cross
                area = cross / 2.0
                if area > best:
                    best = area
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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