<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: int):
        self.root = [i for i in range(n)]
        self.rank = [1] * n
        
    def find(self, x):
        if x == self.root[x]:
            return x
        self.root[x] = self.find(self.root[x])
        return self.root[x]
    
    def union(self, x, y):
        root_x = self.find(x)
        root_y = self.find(y)
        if root_x == root_y:
            return False
        
        if root_x != root_y:
            if self.rank[root_x] > self.rank[root_y]:
                self.root[root_y] = root_x
            elif self.rank[root_y] > self.rank[root_x]:
                self.root[root_x] = root_y
            else:
                self.root[root_y] = root_x
                self.rank[root_x] += 1
        return True
                
    def connected(self, x, y):
        return self.find(x) == self.find(y)
        

class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        n = len(points)
        edges = []
        
        def get_distance(p1, p2):
            return abs(p1[0]-p2[0]) + abs(p1[1]-p2[1])
        
        for i in range(n):
            for j in range(i+1, n):
                distance = get_distance(points[i], points[j])
                edges.append([(i, j), distance])
                
        edges.sort(key=lambda x: x[1])
        
        uf = UnionFind(n)
        
        cost = 0
        num_selected_edges = 0

        for (idx1, idx2), weight in edges:
            if uf.union(idx1, idx2):
                num_selected_edges += 1
                cost += weight
                if num_selected_edges == n-1:
                    break
                
        return cost
            
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