<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 find(self, v):
        if v == self.parent[v]:
            return v
        p = self.find(self.parent[v])
        self.parent[v] = p
        return p
    
    def union(self, a, b):
        a = self.find(a)
        b = self.find(b)

        if self.size[a] < self.size[b]:
            a, b = b,a 
        self.parent[b] = a
        self.size[a] += self.size[b]
        

    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        if len(points) == 1:
            return 0
    
        nodes_added = set()
        from queue import PriorityQueue
        pq = PriorityQueue()
        n = len(points)
        self.parent = [i for i in range(n)]
        self.size = [1 for i in range(n)]
        def get_dist(a, b):
            ans = abs(a[0] - b[0]) + abs(a[1]-b[1])
            return ans

        for i in range(n):
            for j in range(n):
                dist = get_dist(points[i], points[j])
                pq.put((dist, i, j))

        added_count = 0
        cost = 0
        while added_count != len(points)-1:
            e_c, a, b = pq.get()
            if self.find(a) != self.find(b):
                self.union(a, b)
                cost += e_c 
                added_count += 1
        return cost

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