<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 UnionFind():
    def __init__(self,points):
        self.parent = {}
        self.rank = {}

        for i in range(len(points)):
            p1,p2 = points[i]
            self.rank[(p1,p2)] = 0
            self.parent[(p1,p2)] = (p1,p2)

    def find(self,n):
        if n != self.parent[n]:
            self.parent[n] = self.find(self.parent[n])
        return self.parent[n]
    
    def union(self,n1,n2):
        n10,n11 = n1
        n20,n22 = n2
        p1,p2 = self.find((n10,n11)), self.find((n20,n22))
       
        if p1 == p2:
            return False
        if self.rank[p1] > self.rank[p2]:
            self.parent[p2] = p1
        elif self.rank[p1] < self.rank[p2]:
            self.parent[p1] = p2
        else:
            self.parent[p1] = p2
            self.rank[p2] += 1
        return True

def manHatten(x1,x2,y1,y2):
    return abs(x1 -x2) + abs(y1 - y2)
class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        
        minHeap = []
        #get all the costs
        for i in range(len(points)):
            for j in range(i+1,len(points)):
                cost = manHatten(points[i][0], points[j][0], points[i][1], points[j][1])
                heapq.heappush(minHeap,((cost,points[i],points[j])))
        total = 0
        nodes = UnionFind(points)
        
        while minHeap:

            cost,p1,p2= heapq.heappop(minHeap)
            
            if not nodes.union(p1,p2): 
                continue
            else:
                total += cost
        return total



        






        
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