1584. Min Cost to Connect All Points

You are given an array points representing integer coordinates of some points on a 2D-plane, where points[i] = [xi, yi].

The cost of connecting two points [xi, yi] and [xj, yj] is the manhattan distance between them: |xi - xj| + |yi - yj|, where |val| denotes the absolute value of val.

Return the minimum cost to make all points connected. All points are connected if there is exactly one simple path between any two points.

Example 1:

Input: points = [[0,0],[2,2],[3,10],[5,2],[7,0]]
Output: 20
Explanation:

We can connect the points as shown above to get the minimum cost of 20.
Notice that there is a unique path between every pair of points.

ex1

ex2

Example 2:

Input: points = [[3,12],[-2,5],[-4,1]]
Output: 18
Example 3:

Input: points = [[0,0],[1,1],[1,0],[-1,1]]
Output: 4
Example 4:

Input: points = [[-1000000,-1000000],[1000000,1000000]]
Output: 4000000
Example 5:

Input: points = [[0,0]]
Output: 0

Constraints:

  • 1 <= points.length <= 1000
  • -106 <= xi, yi <= 106
  • All pairs (xi, yi) are distinct.

Solution

class Solution {
    int[] a;
    public int minCostConnectPoints(int[][] points) {
        int n = points.length;
        a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = i;
        }
        PriorityQueue<Edge> pq = new PriorityQueue<>();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i != j)  {
                    int[] a = points[i];
                    int[] b = points[j];
                    int w = Math.abs(a[0] - b[0]) + Math.abs(a[1] - b[1]);
                    pq.add(new Edge(i, j, w));
                }
            }
        }
        
        int sum = 0;
        while (pq.size() > 0) {
            Edge e = pq.poll();
            if (!same(e.a, e.b)) {
                union(e.a, e.b);
                sum += e.w;
            }
        }        
        return sum;
    }
    
    boolean same(int p, int q) {
        return find(p) == find(q);
    }
    
    int find(int p) {
        while (a[p] != p) {
            p = a[p];
            a[p] = a[a[p]];
        }
        return p;
    }
    
    void union(int p, int q) {
        int pid = find(p);
        int qid = find(q);
        
        a[qid] = pid;
    }
    
    class Edge implements Comparable<Edge> {
        int a;
        int b;
        int w;
        
        public Edge(int a, int b, int w) {
            this.a = a;
            this.b = b;
            this.w = w;
        }
        
        public int compareTo(Edge other) {
            return Integer.compare(this.w, other.w);
        }
    }
}