<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 UnionFind:
    def __init__(self, n):
        self.p = [i for i in range(n)]

    def find(self, x):
        root = x
        while root != self.p[root]:
            root = self.p[root]
        
        while root != x:
            next_node = self.p[x]
            self.p[x] = root
            x = next_node
        return root
    
    def union(self, node1, node2):
        root1, root2 = self.find(node1), self.find(node2)
        if root1 != root2:
            self.p[root1] = root2
            return True
        return False

def mahattan_dist(x1,y1,x2,y2):
    return abs(x1 - x2) + abs(y1 - y2)

class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        import heapq
        from collections import defaultdict

        # Solution 1: Prims's algorithm
        # graph = defaultdict(dict)
        # for x1,y1 in points:
        #     for x2,y2 in points:
        #         if x1 != x2 or y1 != y2:
        #             val = abs(x1 - x2) + abs(y1 - y2)
        #             graph[(x1,y1)][(x2,y2)] = val
        #             graph[(x2,y2)][(x1,y1)] = val
        
        # visit = {}
        # minHeap = [(0, (points[0][0], points[0][1]))]
        # total_cost = 0
        # while minHeap:
        #     cost, point = heapq.heappop(minHeap)
        #     if not visit.get(point, False):
        #         visit[point] = True
        #         total_cost += cost

        #         for adj_point, adj_cost in graph[point].items():
        #             if not visit.get(adj_point, False):
        #                 heapq.heappush(minHeap, (adj_cost, adj_point))
        # return total_cost

        # Solution 2: Kruskal's algorithm
        minHeap = []
        n = len(points)
        for i in range(n):
            for j in range(n):
                if i != j:
                    val = mahattan_dist(points[i][0],points[i][1],
                                        points[j][0],points[j][1])
                    heapq.heappush(minHeap, (val, (i,j)))
        cost = 0
        e = 0
        uf = UnionFind(n)
        while minHeap:
            w, (i, j) = heapq.heappop(minHeap)
            if uf.union(i,j):
                cost += w
                e += 1
            if e == n - 1:
                break

        return cost
208901