LeetCode #836 — EASY

Rectangle Overlap

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

Problem Statement

An axis-aligned rectangle is represented as a list [x1, y1, x2, y2], where (x1, y1) is the coordinate of its bottom-left corner, and (x2, y2) is the coordinate of its top-right corner. Its top and bottom edges are parallel to the X-axis, and its left and right edges are parallel to the Y-axis.

Two rectangles overlap if the area of their intersection is positive. To be clear, two rectangles that only touch at the corner or edges do not overlap.

Given two axis-aligned rectangles rec1 and rec2, return true if they overlap, otherwise return false.

Example 1:

Input: rec1 = [0,0,2,2], rec2 = [1,1,3,3]
Output: true

Example 2:

Input: rec1 = [0,0,1,1], rec2 = [1,0,2,1]
Output: false

Example 3:

Input: rec1 = [0,0,1,1], rec2 = [2,2,3,3]
Output: false

Constraints:

rec1.length == 4
rec2.length == 4
-10⁹ <= rec1[i], rec2[i] <= 10⁹
rec1 and rec2 represent a valid rectangle with a non-zero area.

Step 01

Brute Force Baseline

Problem summary: An axis-aligned rectangle is represented as a list [x1, y1, x2, y2], where (x1, y1) is the coordinate of its bottom-left corner, and (x2, y2) is the coordinate of its top-right corner. Its top and bottom edges are parallel to the X-axis, and its left and right edges are parallel to the Y-axis. Two rectangles overlap if the area of their intersection is positive. To be clear, two rectangles that only touch at the corner or edges do not overlap. Given two axis-aligned rectangles rec1 and rec2, return true if they overlap, otherwise return false.

Baseline thinking

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

Pattern signal: Math

Example 1

[0,0,2,2]
[1,1,3,3]

Example 2

[0,0,1,1]
[1,0,2,1]

Example 3

[0,0,1,1]
[2,2,3,3]

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

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 #836: Rectangle Overlap
class Solution {
    public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
        int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
        return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
    }
}

// Accepted solution for LeetCode #836: Rectangle Overlap
func isRectangleOverlap(rec1 []int, rec2 []int) bool {
	x1, y1, x2, y2 := rec1[0], rec1[1], rec1[2], rec1[3]
	x3, y3, x4, y4 := rec2[0], rec2[1], rec2[2], rec2[3]
	return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1)
}

# Accepted solution for LeetCode #836: Rectangle Overlap
class Solution:
    def isRectangleOverlap(self, rec1: List[int], rec2: List[int]) -> bool:
        x1, y1, x2, y2 = rec1
        x3, y3, x4, y4 = rec2
        return not (y3 >= y2 or y4 <= y1 or x3 >= x2 or x4 <= x1)

// Accepted solution for LeetCode #836: Rectangle Overlap
struct Solution;

impl Solution {
    fn is_rectangle_overlap(rec1: Vec<i32>, rec2: Vec<i32>) -> bool {
        rec1[2] > rec2[0] && rec1[0] < rec2[2] && rec1[1] < rec2[3] && rec1[3] > rec2[1]
    }
}

#[test]
fn test() {
    let rec1 = vec![0, 0, 2, 2];
    let rec2 = vec![1, 1, 3, 3];
    assert_eq!(Solution::is_rectangle_overlap(rec1, rec2), true);
    let rec1 = vec![0, 0, 1, 1];
    let rec2 = vec![1, 0, 2, 1];
    assert_eq!(Solution::is_rectangle_overlap(rec1, rec2), false);
}

// Accepted solution for LeetCode #836: Rectangle Overlap
function isRectangleOverlap(rec1: number[], rec2: number[]): boolean {
    const [x1, y1, x2, y2] = rec1;
    const [x3, y3, x4, y4] = rec2;
    return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
}

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(1)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.