<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 heapq


class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:


        # min cost to collect all points to going to be the MST 
        # MST := the cheapest tree to connect all nodes without cycles 
        # we know the all edges here are going to positive -> so we're going to want a tree not a cyclic graph 
        # going to have to compute all edges 
        # then do a union find based approach 
        # going to be O(E + Elog(V) + E) -> kruskal's based solution 

        # prim's based solution
        # start at some point -> heap based approach to go to the next point. 
        # O(VlogV) -> for each edge pop
        tree = set(tuple(points[0]))
        res = 0
        fringe = [(0, tuple(points[0]))] 
        while fringe:
            (cost, point) = heapq.heappop(fringe)
            x, y = point
            if point in tree:
                continue
            tree.add(point)
            res += cost
            for p1, p2 in points:
                if point != (p1, p2) and (p1, p2) not in tree:
                    cost = abs(x - p1) + abs(y - p2)
                    heapq.heappush(fringe, (cost, (p1, p2)))

        return res



        
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