<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 UnionFind:
#     def __init__(self, n):
#         self.parent = [*range(n)]
#         self.rank = [1] * n
    
#     def find(self, n):
#         if self.parent[n] != n:
#             self.parent[n] = self.find(self.parent[n])
#         return self.parent[n]
        
#     def union(self, a, b):
#         a, b = self.find(a), self.find(b)
#         if a == b:
#             return
#         if self.rank[a] < self.rank[b]:
#             self.parent[a] = b
#         else:
#             self.parent[b] = a
#             if self.rank[a] == self.rank[b]:
#                 self.rank[a] += 1
#         return


class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        if len(points) == 1:
            return 0
        heap, visited, total = [(0, 1)], [0] * len(points), 0
        min_dist = [inf] * len(points)
        while heap:
            dist, node = heappop(heap)
            if visited[node]:
                continue
            total += dist
            visited[node] = 1
            x1, y1 = points[node][0], points[node][1]
            for neighbor, (x2, y2) in enumerate(points):
                
                dist = abs(x1 - x2) + abs(y1 - y2)
                if dist < min_dist[neighbor]:
                    min_dist[neighbor] = dist
                    if not visited[neighbor]:
                        heappush(heap, (dist, neighbor))
            # print(node, min_dist)
        return total






        # edges = [(u, v, abs(x1-x2) + abs(y1-y2)) for (u, (x1, y1)), (v, (x2, y2)) in combinations(enumerate(points), 2)]
        # uf, total = UnionFind(len(points)), 0
        # for u, v, dist in sorted(edges, key = itemgetter(2)):
        #     if uf.find(u) != uf.find(v):
        #         uf.union(u, v)
        #         total += dist
        # return total
22884