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package g2001_2100.s2013_detect_squares;

// #Medium #Array #Hash_Table #Design #Counting
// #2022_05_24_Time_67_ms_(88.46%)_Space_46.1_MB_(96.15%)

import java.util.HashMap;
import java.util.Map;

/**
 * 2013 - Detect Squares\.
 *
 * Medium
 *
 * 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:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/09/01/image.png)
 *
 * **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`.
**/
public class DetectSquares {
    private static final int MUL = 1002;
    private final Map 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 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;
    }
}




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