<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 len(points)==1:
            return 0
        distance = defaultdict(list)
        for i in range(len(points)):
            for j in range(i+1,len(points)):
                x1 = points[i][0]
                y1 = points[i][1]
                x2 = points[j][0]
                y2 = points[j][1]
                distance[abs(x1-x2)+abs(y1-y2)].append([i,j])
        uf = UnionFind(len(points))
        weights = list(distance.keys())
        weights.sort()
        curr = no_of_edges = 0
        for i in weights :
            for j in distance[i]:
                if uf.find(j[0])!=uf.find(j[1]):
                    curr += i
                    no_of_edges += 1
                    uf.union(j[0],j[1])
                if no_of_edges == len(points)-1:
                    return curr
                
class UnionFind:
    def __init__(self,n):
        self.root = list(range(n))
        self.rank = [1]*n
    def find(self,v):
        if v == self.root[v]:
            return v
        self.root[v] = self.find(self.root[v])
        return self.root[v]
    def union(self,v1,v2):
        rootx = self.find(v1)
        rooty = self.find(v2)
        if rootx!=rooty:
            if self.rank[rootx]>self.rank[rooty]:
                self.root[rooty] = rootx
            elif self.rank[rooty]>self.rank[rootx]:
                self.root[rootx] = rooty
            else:
                self.root[rooty] = rootx
                self.rank[rootx] += 1
    
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