<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 UnionSearch:
    def __init__(self, nodes):
        self.par = {n: n for n in nodes}
        self.rank = {n : 1 for n in nodes}
    
    def search(self, n):
        while n != self.par[n]:
            n = self.par[n]
            self.par[n] = self.par[self.par[n]]
        return n
    
    def union(self, n1, n2):
        p1, p2 = self.search(n1), self.search(n2)
        if p1 == p2:
            return False

        if self.rank[p1] >= self.rank[p2]:
            self.par[p2] = p1
            self.rank[p1] += self.rank[p2]
        else:
            self.par[p1] = p2
            self.rank[p2] += self.rank[p1]
        return True

class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        # apply kruskals algo by finding cost of each edge (dist) between points and then using Union search + heap, for kruskal to construct the mst (what is required) and return the associated cost
        points = [tuple(p) for p in points]
        edges = [] # (weight, p1, p2)
        for i, p1 in enumerate(points):
            for j, p2 in enumerate(points):
                if i == j:
                    continue
                x1, y1 = p1
                x2, y2 = p2
        
                weight = abs(x1 - x2) + abs(y1 - y2)
                edges.append([weight, p1, p2])
        
        heapq.heapify(edges)
        totalWeight = 0
        mstCount = 0 # since mst not needed
        dsu = UnionSearch(points)

        while mstCount < len(points) - 1:
            w, a, b = heapq.heappop(edges)
            if dsu.union(a, b):
                totalWeight += w
                mstCount += 1
        
        return totalWeight
            
            


        

        
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