<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:
        import heapq
        vertices = list(range(len(points)))
        dsu = DSU(vertices)
        dist = lambda x0, y0, x1, y1: abs(x0 - x1) + abs(y0 - y1)
        edges = []
        
        for idx in range(len(points)):
            for jdx in range(idx+1, len(points)):
                xi, yi = points[idx]
                xj, yj = points[jdx]
                cost = dist(xi, yi, xj, yj)
                edges.append((idx, jdx, cost))
        
        edges.sort(key = lambda x: x[2])
        total_cost = 0
        for idx, jdx, cost in edges:
            if dsu.union(idx, jdx):
                total_cost += cost
        
        return total_cost
    
    def prims(self, points):
        from collections import defaultdict, heapq

        dist = lambda x0, y0, x1, y1: abs(x0 - x1) + abs(y0 - y1)
        edges = []
        
        for idx in range(len(points)):
            for jdx in range(idx+1, len(points)):
                xi, yi = points[idx]
                xj, yj = points[jdx]
                cost = dist(xi, yi, xj, yj)
                g[idx].append((cost, jdx))
        
        i, seen = 0, set([0])
        heap = heapq.heapify(g[0])
        
        while len(seen) < len(points):
            cost, jdx = heapq.heappop(heap)
            if jdx not in seen:
                seen.add(jdx)
                total_cost += cost
                for pair in g[jdx]:
                    heapq.heappush(pair)

        return total_cost
        

    
class DSU:
    def __init__(self, vertices):
        self.root = {i: i for i in vertices}
    
    def find(self, x):
        if x != self.root[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:
            self.root[root_y] = root_x
            return True
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