<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>

class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        def getCost(p1_index, p2_index):
            p1, p2 = points[p1_index], points[p2_index]
            return abs(p2[1]-p1[1]) + abs(p2[0]-p1[0])

        edge_heap = []
        for index in range(1, len(points)):
            heapq.heappush(edge_heap, Edge(0, index, getCost(0, index)))
        
        visited = set([0]) 
        min_cost_to_connect_points = 0
        while edge_heap and len(visited) < len(points):
            next_edge = heapq.heappop(edge_heap)
            start, end = next_edge.p1_index, next_edge.p2_index
            if end not in visited:
                visited.add(end)
                min_cost_to_connect_points += next_edge.cost
                for next_index in range(len(points)):
                    if next_index not in visited:
                        heapq.heappush(edge_heap, Edge(end, next_index, getCost(end, next_index)))
        
        return min_cost_to_connect_points
        
class Edge:
    def __init__(self, p1_index, p2_index, cost):
        self.p1_index = p1_index
        self.p2_index = p2_index
        self.cost = cost
        
    def __lt__(self, other):
        return self.cost < other.cost
    
    
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