LeetCode #3588 — MEDIUM

Find Maximum Area of a Triangle

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 2D array coords of size n x 2, representing the coordinates of n points in an infinite Cartesian plane.

Find twice the maximum area of a triangle with its corners at any three elements from coords, such that at least one side of this triangle is parallel to the x-axis or y-axis. Formally, if the maximum area of such a triangle is A, return 2 * A.

If no such triangle exists, return -1.

Note that a triangle cannot have zero area.

Example 1:

Input: coords = [[1,1],[1,2],[3,2],[3,3]]

Output: 2

Explanation:

The triangle shown in the image has a base 1 and height 2. Hence its area is 1/2 * base * height = 1.

Example 2:

Input: coords = [[1,1],[2,2],[3,3]]

Output: -1

Explanation:

The only possible triangle has corners (1, 1), (2, 2), and (3, 3). None of its sides are parallel to the x-axis or the y-axis.

Constraints:

1 <= n == coords.length <= 10⁵
1 <= coords[i][0], coords[i][1] <= 10⁶
All coords[i] are unique.

Step 01

Brute Force Baseline

Problem summary: You are given a 2D array coords of size n x 2, representing the coordinates of n points in an infinite Cartesian plane. Find twice the maximum area of a triangle with its corners at any three elements from coords, such that at least one side of this triangle is parallel to the x-axis or y-axis. Formally, if the maximum area of such a triangle is A, return 2 * A. If no such triangle exists, return -1. Note that a triangle cannot have zero area.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Math · Greedy

Example 1

[[1,1],[1,2],[3,2],[3,3]]

Example 2

[[1,1],[2,2],[3,3]]

Step 02

Core Insight

What unlocks the optimal approach

<code>area * 2 = base * height</code>
Let the base be parallel to the x‑axis or y‑axis
Sort to find the maximum base for each fixed <code>x</code> (or <code>y</code>), then the maximum height comes from the extreme values of the other coordinate.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3588: Find Maximum Area of a Triangle
class Solution {
    public long maxArea(int[][] coords) {
        long ans = calc(coords);
        for (int[] c : coords) {
            int tmp = c[0];
            c[0] = c[1];
            c[1] = tmp;
        }
        ans = Math.max(ans, calc(coords));
        return ans > 0 ? ans : -1;
    }

    private long calc(int[][] coords) {
        int mn = Integer.MAX_VALUE, mx = 0;
        Map<Integer, Integer> f = new HashMap<>();
        Map<Integer, Integer> g = new HashMap<>();

        for (int[] c : coords) {
            int x = c[0], y = c[1];
            mn = Math.min(mn, x);
            mx = Math.max(mx, x);
            if (f.containsKey(x)) {
                f.put(x, Math.min(f.get(x), y));
                g.put(x, Math.max(g.get(x), y));
            } else {
                f.put(x, y);
                g.put(x, y);
            }
        }

        long ans = 0;
        for (var e : f.entrySet()) {
            int x = e.getKey();
            int y = e.getValue();
            int d = g.get(x) - y;
            ans = Math.max(ans, (long) d * Math.max(mx - x, x - mn));
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3588: Find Maximum Area of a Triangle
func maxArea(coords [][]int) int64 {
	calc := func() int64 {
		mn, mx := int(1e9), 0
		f := make(map[int]int)
		g := make(map[int]int)
		for _, c := range coords {
			x, y := c[0], c[1]
			mn = min(mn, x)
			mx = max(mx, x)
			if _, ok := f[x]; ok {
				f[x] = min(f[x], y)
				g[x] = max(g[x], y)
			} else {
				f[x] = y
				g[x] = y
			}
		}
		var ans int64
		for x, y := range f {
			d := g[x] - y
			ans = max(ans, int64(d)*int64(max(mx-x, x-mn)))
		}
		return ans
	}

	ans := calc()
	for _, c := range coords {
		c[0], c[1] = c[1], c[0]
	}
	ans = max(ans, calc())
	if ans > 0 {
		return ans
	}
	return -1
}

# Accepted solution for LeetCode #3588: Find Maximum Area of a Triangle
class Solution:
    def maxArea(self, coords: List[List[int]]) -> int:
        def calc() -> int:
            mn, mx = inf, 0
            f = {}
            g = {}
            for x, y in coords:
                mn = min(mn, x)
                mx = max(mx, x)
                if x in f:
                    f[x] = min(f[x], y)
                    g[x] = max(g[x], y)
                else:
                    f[x] = g[x] = y
            ans = 0
            for x, y in f.items():
                d = g[x] - y
                ans = max(ans, d * max(mx - x, x - mn))
            return ans

        ans = calc()
        for c in coords:
            c[0], c[1] = c[1], c[0]
        ans = max(ans, calc())
        return ans if ans else -1

// Accepted solution for LeetCode #3588: Find Maximum Area of a Triangle
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3588: Find Maximum Area of a Triangle
// class Solution {
//     public long maxArea(int[][] coords) {
//         long ans = calc(coords);
//         for (int[] c : coords) {
//             int tmp = c[0];
//             c[0] = c[1];
//             c[1] = tmp;
//         }
//         ans = Math.max(ans, calc(coords));
//         return ans > 0 ? ans : -1;
//     }
// 
//     private long calc(int[][] coords) {
//         int mn = Integer.MAX_VALUE, mx = 0;
//         Map<Integer, Integer> f = new HashMap<>();
//         Map<Integer, Integer> g = new HashMap<>();
// 
//         for (int[] c : coords) {
//             int x = c[0], y = c[1];
//             mn = Math.min(mn, x);
//             mx = Math.max(mx, x);
//             if (f.containsKey(x)) {
//                 f.put(x, Math.min(f.get(x), y));
//                 g.put(x, Math.max(g.get(x), y));
//             } else {
//                 f.put(x, y);
//                 g.put(x, y);
//             }
//         }
// 
//         long ans = 0;
//         for (var e : f.entrySet()) {
//             int x = e.getKey();
//             int y = e.getValue();
//             int d = g.get(x) - y;
//             ans = Math.max(ans, (long) d * Math.max(mx - x, x - mn));
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3588: Find Maximum Area of a Triangle
function maxArea(coords: number[][]): number {
    function calc(): number {
        let [mn, mx] = [Infinity, 0];
        const f = new Map<number, number>();
        const g = new Map<number, number>();

        for (const [x, y] of coords) {
            mn = Math.min(mn, x);
            mx = Math.max(mx, x);
            if (f.has(x)) {
                f.set(x, Math.min(f.get(x)!, y));
                g.set(x, Math.max(g.get(x)!, y));
            } else {
                f.set(x, y);
                g.set(x, y);
            }
        }

        let ans = 0;
        for (const [x, y] of f) {
            const d = g.get(x)! - y;
            ans = Math.max(ans, d * Math.max(mx - x, x - mn));
        }
        return ans;
    }

    let ans = calc();
    for (const c of coords) {
        [c[0], c[1]] = [c[1], c[0]];
    }
    ans = Math.max(ans, calc());
    return ans > 0 ? ans : -1;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.