LeetCode #391 — HARD

Perfect Rectangle

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

Given an array rectangles where rectangles[i] = [xi, yi, ai, bi] represents an axis-aligned rectangle. The bottom-left point of the rectangle is (xi, yi) and the top-right point of it is (ai, bi).

Return true if all the rectangles together form an exact cover of a rectangular region.

Example 1:

Input: rectangles = [[1,1,3,3],[3,1,4,2],[3,2,4,4],[1,3,2,4],[2,3,3,4]]
Output: true
Explanation: All 5 rectangles together form an exact cover of a rectangular region.

Example 2:

Input: rectangles = [[1,1,2,3],[1,3,2,4],[3,1,4,2],[3,2,4,4]]
Output: false
Explanation: Because there is a gap between the two rectangular regions.

Example 3:

Input: rectangles = [[1,1,3,3],[3,1,4,2],[1,3,2,4],[2,2,4,4]]
Output: false
Explanation: Because two of the rectangles overlap with each other.

Constraints:

  • 1 <= rectangles.length <= 2 * 104
  • rectangles[i].length == 4
  • -105 <= xi < ai <= 105
  • -105 <= yi < bi <= 105

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Given an array rectangles where rectangles[i] = [xi, yi, ai, bi] represents an axis-aligned rectangle. The bottom-left point of the rectangle is (xi, yi) and the top-right point of it is (ai, bi). Return true if all the rectangles together form an exact cover of a rectangular region.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math

Example 1

[[1,1,3,3],[3,1,4,2],[3,2,4,4],[1,3,2,4],[2,3,3,4]]

Example 2

[[1,1,2,3],[1,3,2,4],[3,1,4,2],[3,2,4,4]]

Example 3

[[1,1,3,3],[3,1,4,2],[1,3,2,4],[2,2,4,4]]
Step 02

Core Insight

What unlocks the optimal approach

  • No official hints in dataset. Start from constraints and look for a monotonic or reusable state.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #391: Perfect Rectangle
class Solution {
    public boolean isRectangleCover(int[][] rectangles) {
        long area = 0;
        int minX = rectangles[0][0], minY = rectangles[0][1];
        int maxX = rectangles[0][2], maxY = rectangles[0][3];
        Map<Pair, Integer> cnt = new HashMap<>();

        for (int[] r : rectangles) {
            area += (r[2] - r[0]) * (r[3] - r[1]);

            minX = Math.min(minX, r[0]);
            minY = Math.min(minY, r[1]);
            maxX = Math.max(maxX, r[2]);
            maxY = Math.max(maxY, r[3]);

            cnt.merge(new Pair(r[0], r[1]), 1, Integer::sum);
            cnt.merge(new Pair(r[0], r[3]), 1, Integer::sum);
            cnt.merge(new Pair(r[2], r[3]), 1, Integer::sum);
            cnt.merge(new Pair(r[2], r[1]), 1, Integer::sum);
        }

        if (area != (long) (maxX - minX) * (maxY - minY)
            || cnt.getOrDefault(new Pair(minX, minY), 0) != 1
            || cnt.getOrDefault(new Pair(minX, maxY), 0) != 1
            || cnt.getOrDefault(new Pair(maxX, maxY), 0) != 1
            || cnt.getOrDefault(new Pair(maxX, minY), 0) != 1) {
            return false;
        }

        cnt.remove(new Pair(minX, minY));
        cnt.remove(new Pair(minX, maxY));
        cnt.remove(new Pair(maxX, maxY));
        cnt.remove(new Pair(maxX, minY));

        return cnt.values().stream().allMatch(c -> c == 2 || c == 4);
    }

    private static class Pair {
        final int first;
        final int second;

        Pair(int first, int second) {
            this.first = first;
            this.second = second;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) {
                return true;
            }
            if (o == null || getClass() != o.getClass()) {
                return false;
            }
            Pair pair = (Pair) o;
            return first == pair.first && second == pair.second;
        }

        @Override
        public int hashCode() {
            return Objects.hash(first, second);
        }
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
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.

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