<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:
        if not points or len(points) <= 1:
            return 0

        size = len(points)
        visited = [False] * size
        unvisited = {i for i in range(size)}
        edges = []
        result = 0

        x = points[0][0]
        y = points[0][1]
        for i in range(1, size):
            newX, newY = points[i]
            cost = abs(newX - x) + abs(newY - y)
            edge = Edge(0, i, cost)
            edges.append(edge)

        visited[0] = True
        unvisited.remove(0)

        count = size - 1
        heapq.heapify(edges)
        while edges and count > 0:
            edge = heapq.heappop(edges)
            point1, point2, cost = edge.point1, edge.point2, edge.cost
            if not visited[point2]:
                visited[point2] = True
                unvisited.remove(point2)
                result += cost

                for j in unvisited:
                    x, y = points[point2]
                    newX, newY = points[j]

                    cost = abs(newX - x) + abs(newY - y)
                    heapq.heappush(edges, Edge(point2, j, cost))
                count -= 1

        return result


class Edge:
    def __init__(self, point1, point2, cost):
        self.point1 = point1
        self.point2 = point2
        self.cost = cost

    def __lt__(self, other):
        return self.cost < other.cost




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