2013. Detect Squares

Problem

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.

Solution (Java)

class DetectSquares {
    private static final int MUL = 1002;
    private final Map<Integer, int[]> map;

    public DetectSquares() {
        this.map = new HashMap<>();
    }

    public void add(int[] point) {
        int x = point[0];
        int y = point[1];
        int hash = x * MUL + y;
        if (map.containsKey(hash)) {
            map.get(hash)[2]++;
        } else {
            map.put(hash, new int[] {x, y, 1});
        }
    }

    public int count(int[] point) {
        int ans = 0;
        int x = point[0];
        int y = point[1];
        for (Map.Entry<Integer, int[]> entry : map.entrySet()) {
            int[] diap = entry.getValue();
            int x1 = diap[0];
            int y1 = diap[1];
            int num = diap[2];
            if (Math.abs(x - x1) == Math.abs(y - y1) && x != x1 && y != y1) {
                int p1hash = x * MUL + y1;
                int p2hash = x1 * MUL + y;
                if (map.containsKey(p1hash) && map.containsKey(p2hash)) {
                    ans += map.get(p1hash)[2] * map.get(p2hash)[2] * num;
                }
            }
        }
        return ans;
    }
}

/**
 * Your DetectSquares object will be instantiated and called as such:
 * DetectSquares obj = new DetectSquares();
 * obj.add(point);
 * int param_2 = obj.count(point);
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).