2013. Detect Squares
Description
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 pointpoint = [x, y]
to the data structure.int count(int[] point)
Counts the number of ways to form axis-aligned squares with pointpoint = [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 toadd
andcount
.
Solutions
Solution 1: Hash Table
We can use a hash table $cnt$ to maintain all the information of the points, where $cnt[x][y]$ represents the count of point $(x, y)$.
When calling the $add(x, y)$ method, we increase the value of $cnt[x][y]$ by $1$.
When calling the $count(x_1, y_1)$ method, we need to get three other points to form an axis-aligned square. We can enumerate the point $(x_2, y_1)$ that is parallel to the $x$-axis and at a distance $d$ from $(x_1, y_1)$. If such a point exists, based on these two points, we can determine the other two points as $(x_1, y_1 + d)$ and $(x_2, y_1 + d)$, or $(x_1, y_1 - d)$ and $(x_2, y_1 - d)$. We can add up the number of schemes for these two situations.
In terms of time complexity, the time complexity of calling the $add(x, y)$ method is $O(1)$, and the time complexity of calling the $count(x_1, y_1)$ method is $O(n)$; the space complexity is $O(n)$. Here, $n$ is the number of points in the data stream.
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