Min Cost to Connect All Points

Problem

https://leetcode.com/problems/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] = [x:sub:`i`, y:sub:`i`].

The cost of connecting two points [x:sub:`i`, y:sub:`i`] and [x:sub:`j`, y:sub:`j`] is the manhattan distance between them: |x:sub:`i`- x:sub:`j`| + |y:sub:`i`- y:sub:`j`|, 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

• -10:sup:`6`<= x:sub:`i`, y:sub:`i`<= 10:sup:`6`

• All pairs (x:sub:`i`, y:sub:`i`) are distinct.

Pattern

Array, Union-Find, Graph Theory, Minimum Spanning Tree

Approaches

Prim’s Algorithm

Code

import heapq


def dist(p, q):
    return abs(p[0] - q[0]) + abs(p[1] - q[1])


def minCostConnectPoints(points: list[list[int]]) -> int:
    points = [(x, y) for [x, y] in points]

    visited = set()
    p = points[0]
    visited.add(p)
    edges = [(dist(p, q), p, q) for q in points if p != q]
    heapq.heapify(edges)

    total_cost = 0
    while len(visited) < len(points):
        cost, p, q = heapq.heappop(edges)
        if q in visited:
            continue

        visited.add(q)

        total_cost += cost

        for r in points:
            if r not in visited:
                heapq.heappush(edges, (dist(q, r), q, r))

    return total_cost

Test

>>> from min_cost_to_connect_all_points__prims_algorithm import minCostConnectPoints
>>> minCostConnectPoints([[0,0],[2,2],[3,10],[5,2],[7,0]])
20
>>> minCostConnectPoints([[3,12],[-2,5],[-4,1]])
18

min_cost_to_connect_all_points__prims_algorithm.dist(p, q)

min_cost_to_connect_all_points__prims_algorithm.minCostConnectPoints(points: list[list[int]]) → int