Problem Description

In this problem, you are tasked with managing a collection of points on the Cartesian coordinate system and determining how many distinct squares can be formed with any given query point as one of the corners. The squares of interest must have sides that are aligned with the X and Y axes, which means their sides are either horizontal or vertical, and no rotation is allowed. Additionally, these squares must have a positive area, so the sides need to have a non-zero length.

The operations you have to implement are:

• Adding new points to a collection, even if they are duplicates of existing points.
• For a given query point, counting the number of ways to select three other points from the collection such that together with the query point, they form a proper axis-aligned square.

You are effectively designing a class named DetectSquares that should support these operations. The class will have an add method to include new points into a data structure and a count method to find the number of squares with the query point.

Intuition

The essence of the solution is to find groups of three points in the current collection of points that, along with a given query point, can create an axis-aligned square. Focusing on the constraints of an axis-aligned square, we understand that for any two points that form a side of such a square, their X-coordinates or Y-coordinates must be the same, spelling out a straight horizontal or vertical line.

To arrive at the solution:

• We need an efficient way to store the points and their frequencies (because duplicates are allowed and treated as different points). A dictionary of counters serves this purpose: a dictionary where each key is an X-coordinate and its value is a counter that keeps track of the number and frequency of Y-coordinates that appear at this X-coordinate.

• Adding a new point involves inserting the point into the dictionary or updating the count if it's already there.

• When counting the number of ways to form a square with a given point as one of its corners, you need to iterate over all points that could potentially be on the corners of squares that can be formed with the query point. Essentially, for each point with the same Y-coordinate as the query point but a different X-coordinate (forming a potential side of a square along the X-axis), check if there are points above and below (or at the opposite side on the same level) that could complete the square.

• This can be visualized by examining the vertical and horizontal lines from the query point, calculating the potential side length of a square, and then looking for points that would form the opposite corners of the square, thus completing it.

• The total count is the sum of all legitimate squares that can be formed with the query point and the points in the structure, considering all possible distances and directions (up and down from the query point).

By implementing a class with these functionalities, we can efficiently track points and count the squares formed with any given query point.

Solution Approach

The algorithm utilizes a combination of hash tables and arithmetic to efficiently solve the problem at hand. Below is a breakdown of the strategy using the various components within the solution code.

Data Structure Selection

A defaultdict of Counter objects is selected to maintain the frequency of points at different coordinates. This nested data structure is optimal because it allows:

• O(1) average time complexity for adding points.
• O(1) average time complexity for lookups.
• The inner Counter keeps track of multiple Y-coordinates associated with a single X-coordinate.

add Method Implementation

When adding a point (x, y), the method updates the counter for the specified X-coordinate with the Y-coordinate. The Counter object efficiently keeps track of the number of times the same point is added.

def add(self, point: List[int]) -> None:
    x, y = point
    self.cnt[x][y] += 1

count Method Implementation

Counting the number of squares is more complex. We execute the following steps:

1. Retrieve the X and Y coordinates of the query point.
2. If the X-coordinate of the query point is absent from our data structure, it cannot form a square, so we return 0.
3. Iterate over all stored X-coordinates, excluding the X-coordinate of the query point.
4. Calculate the potential side length d of the square as the difference between the current X-coordinate and the query's X-coordinate.
5. Count the squares by using the formula:

self.cnt[x2][y1] * self.cnt[x1][y1 + d] * self.cnt[x2][y1 + d]

This formula checks for a point on the same Y-coordinate as the query but at a different X (self.cnt[x2][y1]), and then looks for points that can form the opposite corners of the square (self.cnt[x1][y1 + d] and self.cnt[x2][y1 + d]).

The same calculation is done for a potential square below the query point. This involves a similar calculation, but we subtract d from y1 to form the lower side of the square.

def count(self, point: List[int]) -> int:
    x1, y1 = point
    if x1 not in self.cnt:
        return 0
    ans = 0
    for x2 in self.cnt.keys():
        if x2 != x1:
            d = x2 - x1
            ans += self.cnt[x2][y1] * self.cnt[x1][y1 + d] * self.cnt[x2][y1 + d]
            ans += self.cnt[x2][y1] * self.cnt[x1][y1 - d] * self.cnt[x2][y1 - d]
    return ans

The method iterates through all X-coordinates (except the query's X-coordinate), for each trying both upward and downward distances, and multiplies the corresponding point frequencies, thereby counting the number of valid squares that can be formed at that potential side length.

The total sum ans is accumulated and returned as the final count.

Conclusion

The solution leverages the constant-time lookups of hash tables and combines it with simple coordinate arithmetic to efficiently handle a stream of points and queries. This balance of data structure choice and algorithm ensures scalability with respect to a high number of points and queries.

Example Walkthrough

Consider a DetectSquares class instance. Initially, there are no points in our collection. Let's walk through a small example of adding points and querying the number of possible squares.

1. Assume we add the points (1, 1), (1, 2), and (2, 1) to the collection by calling the add method for each point.

   After adding these points, our data structure looks something like this:

   {
     1: Counter({1: 1, 2: 1}),
     2: Counter({1: 1})
   }

   This reflects that we have points at (1, 1), (1, 2), and (2, 1) with each point occurring exactly once.

2. Now, add another point to the collection: (2, 2). The updated data structure becomes:

   {
     1: Counter({1: 1, 2: 1}),
     2: Counter({1: 1, 2: 1})
   }

   We have added point (2, 2), and our coordinates are symmetric about point (1, 1) and (2, 2).

3. If we query with the point (1, 1) using the count method, we perform the following:

   Retrieve the x and y coordinates of the query point, which are (x1, y1) = (1, 1).

   Iterate over all stored x-coordinates (we have x = 1 and x = 2 in our data structure):

   For x = 2:

   • Calculate the potential side length d as the difference in x-coordinates, which is 2 - 1 = 1.

   • Check for point at (2, 1), which is the point on the same y-coordinate but a different x-coordinate.

   • Check for points at (1, 1 + d) = (1, 2) and (2, 1 + d) = (2, 2), which are potential opposite corners of a square with side length d.

   • The number of squares for this side length can be counted as self.cnt[x2][y1] * self.cnt[x1][y1 + d] * self.cnt[x2][y1 + d] = 1 * 1 * 1 = 1.

   • Since the square must have positive area, we don't consider squares below the query point because (1, 1 - d) = (1, 0) does not exist in our data structure.

   The total number of squares with the query point (1, 1) as a corner is 1.

The described method accurately counts the squares with sides parallel to the x and y axes without rotating the points, considering both potential squares above and below the queried point (if applicable). This makes it efficient and reliable for rapidly answering multiple queries after incremental additions to the point collection.

Solution Implementation

Time and Space Complexity

Time Complexity

• __init__ method: The initialization of the DetectSquares class sets up a default dictionary with a Counter as its default value. The time complexity for this operation is O(1) because it's a simple assignment operation.

• add method: Adds a point with coordinates (x, y). This part has a time complexity of O(1) because it's just incrementing the count of the point (x, y) in a hash table, which is an O(1) operation.

• count method: This calculates the number of squares with one corner at the given point. The method iterates through all unique x coordinates stored in the hash map (the keys of the cnt dictionary). For each x2 other than x1, it calculates two potential square sides (d = x2 - x1 for the side to the right of (x1, y1) and d = x1 - x2 for the side to the left of (x1, y1)). It then checks for points that are d distance away on the y-axis both above and below (x1,y1).

  • For a given point, the number of potential squares is computed by considering each different x2 coordinate as a potential corner. If there are n different x coordinates, there could be up to n - 1 iterations (discounting the current x coordinate).

  • Inside each iteration, the calculation ans += ... involves constant-time dictionary lookups and arithmetic operations. So each iteration can be considered O(1).

  • Therefore, the time complexity for this method will be O(n), where n is the number of unique x coordinates stored so far when count is called.

The time complexity of the add method is insignificant compared to the count method which dominates when analyzing the complexity of the class operations. Hence, the overall time complexity hinges on the count method which is O(n) in the worst case.

Space Complexity

• The space complexity of the DetectSquares class is dominated by the self.cnt dictionary, which stores the counts of points. This dictionary can potentially grow to include every unique point seen so far.

• If the total number of points is p, and there are u unique x-coordinates and v unique y-coordinates amongst them, the maximum storage required would be for a complete mapping of these points, which is O(u * v).

• In the worst-case scenario where every added point has a unique (x, y) combination, the space complexity would be O(p), where p is the number of add operations performed.

Therefore, the space complexity of the DetectSquares class is O(p).

Learn more about how to find time and space complexity quickly using problem constraints.