<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
from typing import List, DefaultDict


class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        # We want the minimum cost to connect all points -> SUM of edges of Minimum Spanning Tree
        # Kruskal or Prims? Let's try both, but I think Prims might require unique nodes or indices, Kruskal might be easier
        # We're used to getting an adjacency list, but given only nodes and how to calculate edges
        # N^2 edges need to be evaluated, not sparse graph -> Prims > Kruskal

        class UnionFind:
            # Where n is the NUMBER of nodes in graph
            def __init__(self, n) -> None:
                self.parent = {}
                self.rank = {}

                for i in range(n):
                    self.parent[i] = i
                    self.rank[i] = 0

            def find(self, n):
                p = self.parent[n]

                while p != self.parent[p]:
                    self.parent[p] = self.parent[self.parent[p]]
                    p = self.parent[p]

                return p

            def union(self, n1, n2):
                p1, p2 = self.find(n1), self.find(n2)

                # They are eachothers parents
                if p1 == p2:
                    return False

                if self.rank[p1] > self.rank[p2]:
                    self.parent[p2] = p1
                elif self.rank[p1] < self.rank[p2]:
                    self.parent[p1] = p2
                else:
                    self.parent[p1] = p2
                    self.rank[p2] += 1
                return True

        # Build adjList
        # n^2 time to build
        n = len(points)
        adjList = DefaultDict(list)

        for i in range(n):
            xi, yi = points[i]
            for j in range(i + 1, n):
                xj, yj = points[j]
                weight = abs(xi - xj) + abs(yi - yj)
                # adjList needs to AT LEAST contain weight, dest node
                adjList[i].append([weight, j])

        minHeap = []
        for n1 in adjList:
            for weight, n2 in adjList[n1]:
                heapq.heappush(minHeap, [weight, n1, n2])

        # Number of nodes
        unionFind = UnionFind(n)
        mst = []
        cost = 0
        # E edges to traverse, edges < vertices - 1 base case
        while len(mst) < n - 1:
            # log(E) => log(V) time to pop
            weight, n1, n2 = heapq.heappop(minHeap)
            # log or Ackerman time for union
            if not unionFind.union(n1, n2):
                continue
            mst.append([n1, n2])
            cost += weight
        return cost  # mst
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