LeetCode #3380 — MEDIUM

Maximum Area Rectangle With Point Constraints I

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

The Problem

Problem Statement

You are given an array points where points[i] = [x_i, y_i] represents the coordinates of a point on an infinite plane.

Your task is to find the maximum area of a rectangle that:

Can be formed using four of these points as its corners.
Does not contain any other point inside or on its border.
Has its edges parallel to the axes.

Return the maximum area that you can obtain or -1 if no such rectangle is possible.

Example 1:

Input: points = [[1,1],[1,3],[3,1],[3,3]]

Output: 4

Explanation:

We can make a rectangle with these 4 points as corners and there is no other point that lies inside or on the border. Hence, the maximum possible area would be 4.

Example 2:

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

Output: -1

Explanation:

There is only one rectangle possible is with points [1,1], [1,3], [3,1] and [3,3] but [2,2] will always lie inside it. Hence, returning -1.

Example 3:

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

Output: 2

Explanation:

The maximum area rectangle is formed by the points [1,3], [1,2], [3,2], [3,3], which has an area of 2. Additionally, the points [1,1], [1,2], [3,1], [3,2] also form a valid rectangle with the same area.

Constraints:

1 <= points.length <= 10
points[i].length == 2
0 <= x_i, y_i <= 100
All the given points are unique.

Step 01

Brute Force Baseline

Problem summary: You are given an array points where points[i] = [xi, yi] represents the coordinates of a point on an infinite plane. Your task is to find the maximum area of a rectangle that: Can be formed using four of these points as its corners. Does not contain any other point inside or on its border. Has its edges parallel to the axes. Return the maximum area that you can obtain or -1 if no such rectangle is possible.

Baseline thinking

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

Pattern signal: Array · Math · Segment Tree

Example 1

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

Example 2

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

Example 3

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

Core Insight

What unlocks the optimal approach

If <code>(x1, y1)</code> and <code>(x2, y2)</code> are two opposite corners of a rectangle, then the other two would be <code>(x1, y2)</code> and <code>(x2, y1)</code>.
Fix two points and find the other two using a set data structure.
After determining the rectangle, iterate through the array of points to ensure no point lies on the rectangle’s border or within its interior.

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 #3380: Maximum Area Rectangle With Point Constraints I
class Solution {
    public int maxRectangleArea(int[][] points) {
        int ans = -1;
        for (int i = 0; i < points.length; ++i) {
            int x1 = points[i][0], y1 = points[i][1];
            for (int j = 0; j < i; ++j) {
                int x2 = points[j][0], y2 = points[j][1];
                int x3 = Math.min(x1, x2), y3 = Math.min(y1, y2);
                int x4 = Math.max(x1, x2), y4 = Math.max(y1, y2);
                if (check(points, x3, y3, x4, y4)) {
                    ans = Math.max(ans, (x4 - x3) * (y4 - y3));
                }
            }
        }
        return ans;
    }

    private boolean check(int[][] points, int x1, int y1, int x2, int y2) {
        int cnt = 0;
        for (var p : points) {
            int x = p[0];
            int y = p[1];
            if (x < x1 || x > x2 || y < y1 || y > y2) {
                continue;
            }
            if ((x == x1 || x == x2) && (y == y1 || y == y2)) {
                cnt++;
                continue;
            }
            return false;
        }
        return cnt == 4;
    }
}

// Accepted solution for LeetCode #3380: Maximum Area Rectangle With Point Constraints I
func maxRectangleArea(points [][]int) int {
	check := func(x1, y1, x2, y2 int) bool {
		cnt := 0
		for _, point := range points {
			x, y := point[0], point[1]
			if x < x1 || x > x2 || y < y1 || y > y2 {
				continue
			}
			if (x == x1 || x == x2) && (y == y1 || y == y2) {
				cnt++
				continue
			}
			return false
		}
		return cnt == 4
	}

	ans := -1
	for i := 0; i < len(points); i++ {
		x1, y1 := points[i][0], points[i][1]
		for j := 0; j < i; j++ {
			x2, y2 := points[j][0], points[j][1]
			x3, y3 := min(x1, x2), min(y1, y2)
			x4, y4 := max(x1, x2), max(y1, y2)
			if check(x3, y3, x4, y4) {
				ans = max(ans, (x4-x3)*(y4-y3))
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #3380: Maximum Area Rectangle With Point Constraints I
class Solution:
    def maxRectangleArea(self, points: List[List[int]]) -> int:
        def check(x1: int, y1: int, x2: int, y2: int) -> bool:
            cnt = 0
            for x, y in points:
                if x < x1 or x > x2 or y < y1 or y > y2:
                    continue
                if (x == x1 or x == x2) and (y == y1 or y == y2):
                    cnt += 1
                    continue
                return False
            return cnt == 4

        ans = -1
        for i, (x1, y1) in enumerate(points):
            for x2, y2 in points[:i]:
                x3, y3 = min(x1, x2), min(y1, y2)
                x4, y4 = max(x1, x2), max(y1, y2)
                if check(x3, y3, x4, y4):
                    ans = max(ans, (x4 - x3) * (y4 - y3))
        return ans

// Accepted solution for LeetCode #3380: Maximum Area Rectangle With Point Constraints I
// 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 #3380: Maximum Area Rectangle With Point Constraints I
// class Solution {
//     public int maxRectangleArea(int[][] points) {
//         int ans = -1;
//         for (int i = 0; i < points.length; ++i) {
//             int x1 = points[i][0], y1 = points[i][1];
//             for (int j = 0; j < i; ++j) {
//                 int x2 = points[j][0], y2 = points[j][1];
//                 int x3 = Math.min(x1, x2), y3 = Math.min(y1, y2);
//                 int x4 = Math.max(x1, x2), y4 = Math.max(y1, y2);
//                 if (check(points, x3, y3, x4, y4)) {
//                     ans = Math.max(ans, (x4 - x3) * (y4 - y3));
//                 }
//             }
//         }
//         return ans;
//     }
// 
//     private boolean check(int[][] points, int x1, int y1, int x2, int y2) {
//         int cnt = 0;
//         for (var p : points) {
//             int x = p[0];
//             int y = p[1];
//             if (x < x1 || x > x2 || y < y1 || y > y2) {
//                 continue;
//             }
//             if ((x == x1 || x == x2) && (y == y1 || y == y2)) {
//                 cnt++;
//                 continue;
//             }
//             return false;
//         }
//         return cnt == 4;
//     }
// }

// Accepted solution for LeetCode #3380: Maximum Area Rectangle With Point Constraints I
function maxRectangleArea(points: number[][]): number {
    const check = (x1: number, y1: number, x2: number, y2: number): boolean => {
        let cnt = 0;
        for (const point of points) {
            const [x, y] = point;
            if (x < x1 || x > x2 || y < y1 || y > y2) {
                continue;
            }
            if ((x === x1 || x === x2) && (y === y1 || y === y2)) {
                cnt++;
                continue;
            }
            return false;
        }
        return cnt === 4;
    };

    let ans = -1;
    for (let i = 0; i < points.length; i++) {
        const [x1, y1] = points[i];
        for (let j = 0; j < i; j++) {
            const [x2, y2] = points[j];
            const [x3, y3] = [Math.min(x1, x2), Math.min(y1, y2)];
            const [x4, y4] = [Math.max(x1, x2), Math.max(y1, y2)];
            if (check(x3, y3, x4, y4)) {
                ans = Math.max(ans, (x4 - x3) * (y4 - y3));
            }
        }
    }
    return ans;
}

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 + q log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n × q) time

O(1) space

For each of q queries, scan the entire range to compute the aggregate (sum, min, max). Each range scan takes up to O(n) for a full-array query. With q queries: O(n × q) total. Point updates are O(1) but queries dominate.

SEGMENT TREE

O(n + q log n) time

O(n) space

Building the tree is O(n). Each query or update traverses O(log n) nodes (tree height). For q queries: O(n + q log n) total. Space is O(4n) ≈ O(n) for the tree array. Lazy propagation adds O(1) per node for deferred updates.

Shortcut: Build O(n), query/update O(log n) each. When you need both range queries AND point updates.

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.

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.

Maximum Area Rectangle With Point Constraints I

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

Overflow in intermediate arithmetic