<p>You are given an array <code>points</code> representing integer coordinates of some points on a 2D-plane, where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>.</p>

<p>The cost of connecting two points <code>[x<sub>i</sub>, y<sub>i</sub>]</code> and <code>[x<sub>j</sub>, y<sub>j</sub>]</code> is the <strong>manhattan distance</strong> between them: <code>|x<sub>i</sub> - x<sub>j</sub>| + |y<sub>i</sub> - y<sub>j</sub>|</code>, where <code>|val|</code> denotes the absolute value of <code>val</code>.</p>

<p>Return <em>the minimum cost to make all points connected.</em> All points are connected if there is <strong>exactly one</strong> simple path between any two points.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/26/d.png" style="width: 214px; height: 268px;" />
<pre>
<strong>Input:</strong> points = [[0,0],[2,2],[3,10],[5,2],[7,0]]
<strong>Output:</strong> 20
<strong>Explanation:</strong> 
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/26/c.png" style="width: 214px; height: 268px;" />
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> points = [[3,12],[-2,5],[-4,1]]
<strong>Output:</strong> 18
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= points.length &lt;= 1000</code></li>
	<li><code>-10<sup>6</sup> &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>6</sup></code></li>
	<li>All pairs <code>(x<sub>i</sub>, y<sub>i</sub>)</code> are distinct.</li>
</ul>

import math
from collections import defaultdict
import heapq
class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        """
        1 - start with a point, on top of queue
        2 - git distance to all points,
                if distnace to some point smaller than before
                update min distance
             add to queue, 
        3 - mark current point as visited
        4 - repeat 1 - 3 while queue

        """

        visit=set()
        pq = []
        heapq.heappush(pq, (0, (points[0])))

        min_distances={(point[0],point[1]):math.inf for point in points}
        min_distances[(points[0][0],points[0][1])]=0
        while pq and len(visit) != len(points):
            min_distance, current_point = heapq.heappop(pq)
            visit.add((current_point[0],current_point[1]))
            for point in points:
                if (point[0],point[1]) not in visit:
                    distance_to_point = abs(current_point[0]-point[0]) + abs(current_point[1]-point[1])
                    if distance_to_point < min_distances[(point[0],point[1])]:
                        min_distances[(point[0],point[1])]=distance_to_point
                        heapq.heappush(pq, (distance_to_point, (point)))
        
        return sum(min_distances.values())


20370