<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:
        parent = {tuple(i):tuple(i) for i in points}
        rank = {tuple(i):1 for i in points}
        res = 0
        minHeap = []
        for i in range(len(points)):
            r1 , c1 = points[i]
            for j in range(i+1,len(points)):
                r2 , c2 = points[j]
                val = abs(abs(r2-r1)+abs(c2-c1))
                heapq.heappush(minHeap,([val,(r1,c1),(r2,c2)]))
        
        def find(node):
            p = parent[node]
            while p != parent[p]:
                parent[p] = parent[parent[p]]
                p = parent[p]
            return p

        def union(n1,n2):
            p1 , p2 = find(n1) , find(n2)
            if p1 == p2 : return False
            if rank[p1] > rank[p2]:
                parent[p2] = p1
            elif rank[p2] > rank[p1]:
                parent[p1] = p2
            else:
                parent[p2] = p1
                rank[p1] += 1  
            return True

        while minHeap :
            w , s , d = heapq.heappop(minHeap)
            if union(s,d):
                res += w
        return res
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