<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:
        n = len(points)
        edges = []
        for i in range(n-1):
            for j in range(i+1, n):
                x1, y1 = points[i]
                x2, y2 = points[j]
                edges.append((abs(x1-x2) + abs(y1-y2), i, j))
        edges.sort()
        count = 0
        mst_cost = 0

        parent = list(range(n))
        rank = [0] * n

        def union(x, y):
            x = find(x)
            y = find(y)
            if rank[x] > rank[y]:
                parent[y] = x
            else:
                parent[x] = y
                if rank[x] == rank[y]:
                    rank[y] += 1

        def find(x):
            if parent[x] != x:
                parent[x] = find(parent[x])
            return parent[x]
            
        for w, u, v in edges:
            if find(u) != find(v):
                mst_cost += w
                union(u, v)
                count += 1
                if count == n - 1:
                    break
        return mst_cost

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