1584. Min Cost to Connect All Points

Problem

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.

Example 2:

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

  Constraints:

• 1 <= points.length <= 1000

• -106 <= xi, yi <= 106

• All pairs (xi, yi) are distinct.

Solution

/**
 * @param {number[][]} points
 * @return {number}
 */
var minCostConnectPoints = function(points) {
    var edges = [];
    for (var i = 0; i < points.length; i++) {
        for (var j = i + 1; j < points.length; j++) {
            edges.push([i, j, getDistance(points, i, j)]);
        }
    }
    edges.sort((a, b) => a[2] - b[2]);
    var res = 0;
    var parents = Array(points.length).fill(0).map((_, i) => i);
    var rank = Array(points.length).fill(0);
    for (var m = 0; m < edges.length; m++) {
        var [i, j, distance] = edges[m];
        var n = find(parents, i);
        var k = find(parents, j);
        if (n === k) continue;
        if (rank[n] <= rank[k]) {
            union(parents, n, k);
            rank[k]++;
        } else {
            union(parents, k, n);
            rank[n]++;
        }
        res += distance;
    }
    return res;
};

var getDistance = function(points, i, j) {
    return Math.abs(points[i][0] - points[j][0]) + Math.abs(points[i][1] - points[j][1]);
};

var union = function(parents, i, j) {
    parents[i] = j;
};

var find = function(parents, i) {
    if (parents[i] === i) {
        return i;
    }
    parents[i] = find(parents, parents[i]);
    return parents[i];
};

Explain:

Union-find to detect circle.

Kruskal’s Algorithm to get minimum spanning tree.

Complexity:

• Time complexity : O(n ^ 2).
• Space complexity : O(n ^ 2).