# Solving Leetcode Interviews in Seconds with AI: Detect Squares


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2013" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given a stream of points on the X-Y plane. Design an algorithm that:  Adds new points from the stream into a data structure. 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 y-axis. Implement the DetectSquares class:  DetectSquares() Initializes the object with an empty data structure. void add(int[] point) Adds a new point point = [x, y] to the data structure. int count(int[] point) Counts the number of ways to form axis-aligned squares with point point = [x, y] as described above.    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 [null, null, null, null, 1, 0, null, 2]  Explanation DetectSquares detectSquares = new DetectSquares(); detectSquares.add([3, 10]); detectSquares.add([11, 2]); detectSquares.add([3, 2]); detectSquares.count([11, 10]); // return 1. You can choose:                                //   - The first, second, and third points detectSquares.count([14, 8]);  // return 0. The query point cannot form a square with any points in the data structure. detectSquares.add([11, 2]);    // Adding duplicate points is allowed. detectSquares.count([11, 10]); // return 2. You can choose:                                //   - The first, second, and third points                                //   - The first, third, and fourth points    Constraints:  point.length == 2 0 <= x, y <= 1000 At most 3000 calls in total will be made to add and count.  

	# Explanation
	Here's a solution to the DetectSquares problem:

*   **High-Level Approach:**
    *   Store points and their frequencies using a dictionary/hash map. The keys will be x-coordinates, and the values will be another dictionary storing y-coordinates and their frequencies.
    *   For the `count` operation, iterate through all points with the same x-coordinate as the query point. Check if a square can be formed, and if so, add the number of ways it can be formed to a counter.
    *   The number of ways can be calculated using the frequencies of points in the hash map, and by checking if other points necessary to form the square exist.

*   **Complexity:**
    *   Runtime: `add` is O(1), `count` is O(N), where N is the number of points with the same x-coordinate as the query point. In the worst-case scenario, the complexity will be O(M), where M is the total number of points added.
    *   Storage: O(M), where M is the number of points added (due to storing point counts in a hash map).

	
	# Code
	```python
	class DetectSquares:

    def __init__(self):
        self.points = {}  # {x: {y: count}}

    def add(self, point):
        x, y = point
        if x not in self.points:
            self.points[x] = {}
        if y not in self.points[x]:
            self.points[x][y] = 0
        self.points[x][y] += 1

    def count(self, point):
        x, y = point
        count = 0
        if x not in self.points:
            return 0

        for y1 in self.points[x]:
            if y1 == y:
                continue

            side = abs(y1 - y)

            # Check for the two other possible points to make square
            x1 = x + side
            if x1 in self.points:
                if y in self.points[x1] and y1 in self.points[x1]:
                    count += self.points[x][y1] * self.points[x1][y] * self.points[x1][y1]

            x1 = x - side
            if x1 in self.points:
                if y in self.points[x1] and y1 in self.points[x1]:
                    count += self.points[x][y1] * self.points[x1][y] * self.points[x1][y1]
        return count
	```