Find the Largest Area of Square Inside Two Rectangles

Find the Largest Area of Square Inside Two Rectangles - Problem

There exist n rectangles in a 2D plane with edges parallel to the x and y axis. You are given two 2D integer arrays bottomLeft and topRight where:

bottomLeft[i] = [a_i, b_i] represents the bottom-left coordinates of the i-th rectangle

topRight[i] = [c_i, d_i] represents the top-right coordinates of the i-th rectangle

You need to find the maximum area of a square that can fit inside the intersecting region of at least two rectangles. Return 0 if such a square does not exist.

Input & Output

Example 1 — Basic Intersection

$ Input: bottomLeft = [[1,1],[2,2],[3,1]], topRight = [[3,3],[4,4],[4,4]]

› Output: 1

💡 Note: Rectangle 0: (1,1) to (3,3), Rectangle 1: (2,2) to (4,4). Their intersection is (2,2) to (3,3) with width=1, height=1. Largest square has side=1, area=1.

Example 2 — Larger Intersection

$ Input: bottomLeft = [[1,1],[2,2],[1,2]], topRight = [[4,4],[4,4],[3,4]]

› Output: 4

💡 Note: Rectangle 0: (1,1) to (4,4), Rectangle 1: (2,2) to (4,4). Intersection is (2,2) to (4,4) with width=2, height=2. Largest square has side=2, area=4.

Example 3 — No Intersection

$ Input: bottomLeft = [[1,1],[3,3]], topRight = [[2,2],[4,4]]

› Output: 0

💡 Note: Rectangle 0: (1,1) to (2,2), Rectangle 1: (3,3) to (4,4). These rectangles don't intersect, so no square can fit in a shared region.

Constraints

2 ≤ n ≤ 5 × 10⁴
bottomLeft[i].length == topRight[i].length == 2
-10⁹ ≤ bottomLeft[i][j], topRight[i][j] ≤ 10⁹
bottomLeft[i][0] < topRight[i][0] and bottomLeft[i][1] < topRight[i][1]

Visualization

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Asked in

G Google 45 a Amazon 32 M Microsoft 28

The key insight is to find all pairwise rectangle intersections and calculate the largest square that fits in each. For each intersection with width W and height H, the largest square has side length min(W,H). Best approach is the optimized O(n²) solution with early termination. Time: O(n²), Space: O(1).

Common Approaches

✓ Stack

⏱️ Time: N/A Space: N/A

Brute Force - Check All Pairs

⏱️ Time: O(n²) Space: O(1)

For each pair of rectangles, calculate their intersection area. If intersection exists, find the largest square that fits inside it. The side of the largest square equals the minimum of intersection width and height.

Optimized Single Pass

⏱️ Time: O(n²) Space: O(1)

Uses the same pair-checking logic but adds optimizations like early termination when no intersection is possible and skipping duplicate calculations.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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