11. Minimum spanning tree - points

MST (Minimum Spanning Tree)

1584. Min Cost to Connect All PointsMedium3.7K85Companies

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.

Min Cost to Connect All Points - LeetCodeLeetCode

Prim

---

Time: Prim: ((V + E)logN), E insert logV ⇒ ElogV, V remove logV ⇒ VlogV adjs Creation: O(N^2) Total: O(N^2)

Space: O(N^2) every points can be connected to each other

/**
 * @param {number[][]} points
 * @return {number}
 */

function Node(vertex, precedessor, dist) {
    this.vertex = vertex;
    this.precedessor = precedessor;
    this.dist = dist;
}

var minCostConnectPoints = function (points) {
    // first need to find weights between each point pair
    const dist = {};

		// V^2
    for (let i = 0; i < points.length; i++) {
        dist[i] = [];
        for (let j = 0; j < points.length; j++) {
            const distance = getDistance(points[i], points[j]);
            dist[i][j] = dist[i][j] ?? [];
            dist[i][j] = distance;
        }
    }

    const minHeap = new Heap([], (a, b) => a.dist <= b.dist);
    
    const visited = new Set();
    minHeap.insert(new Node(0, 0, 0));

    let sum = 0;
		// V
    while (minHeap.size > 0) {
				// logV
        const node = minHeap.remove();

        // required filter
        if (visited.has(node.vertex)) {
            continue;
        }

        sum += node.dist;
        visited.add(node.vertex);

        const adj = dist[node.vertex];
				// E
        for (let there = 0; there < adj.length; there++) {
            // can do another filter optional
            // if (visited.has(there)) {
            //     continue;
            // }
            const d = dist[node.vertex][there];
						// logV
            minHeap.insert(new Node(there, node.vertex, d));
        }
    }   

    return sum;
};

function getDistance(p1, p2) {
    return Math.abs(p1[0] - p2[0]) + Math.abs(p1[1] - p2[1]);
}

function swap(arr, a, b) {
    [arr[a], arr[b]] = [arr[b], arr[a]];
}