What is the optimal time complexity of solving Detect Squares?

The optimized solution achieves a time complexity of O(n) by using sorting, mathematical shortcuts, or hash table lookups.

What is the auxiliary space complexity for Detect Squares?

The space complexity is O(n) since we only use a minimal/constant amount of extra memory for tracking variables (or auxiliary space if dynamic structures are allocated).

What are the typical edge cases we must handle in Detect Squares?

Key edge cases include empty collections, negative or large bounds causing integer overflow, arrays with only one element, or inputs where target criteria are not met.

How can we optimize the brute force approach for Detect Squares?

Brute force methods often inspect all possible combinations. We optimize this by sorting the input list to enable binary search, or tracking processed items in a set to avoid redundant nested loop lookups.

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Detect Squares

Learn how to solve the 'Detect Squares' problem. This detailed resource details brute force and optimized approaches.

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Problem Statement

Easy

You are given a stream of points on the X-Y plane. Design a data structure that:

- Adds new points from the stream. Duplicate points are allowed and should be treated as different points.

- Given a query point, counts the number of ways to choose three points from the data structure such that the three points and the query point form an axis-aligned square with positive area.

An axis-aligned square is a square whose edges are all the same length and are either parallel or perpendicular to the x-axis and the y-axis.

Implement a function detectSquares(operations: list, arguments: list) -> list where operations are 'DetectSquares', 'add', or 'count', and arguments are the corresponding parameters. Returns a list of results (None for constructor and add).

Constraints

•point.length == 2
•0 <= x, y <= 1000
•At most 3000 calls in total will be made to add and count

Examples

Example 1

Input

["DetectSquares","add","add","add","count","count","add","count"], [[],[3,10],[11,2],[3,2],[11,10],[14,8],[11,2],[11,10]]

Output

[None,None,None,None,1,0,None,2]

Explanation

After adding (3,10), (11,2), (3,2): count(11,10) finds 1 square with corners (3,10),(11,10),(11,2),(3,2). count(14,8) finds 0. After adding another (11,2): count(11,10) finds 2 squares (using each copy of (11,2)).

Need a Hint?

Analyze the input constraints. Try sorting first (O(n log n)) or using a hash map/set to track seen elements in O(n) time.

Edge Cases to Watch

Empty list or null input variables
Single item lists/arrays
Extremely large input bounds causing integer or stack overflow

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