<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 Edge:
    def __init__(self,point1,point2,cost):
        self.point1 = point1
        self.point2 = point2
        self.cost = cost
        
    def __lt__(self,other):
        return self.cost <= other.cost
    
class UnionFind:
    def __init__(self, size):
        self.root = [i for i in range(size)]
        self.rank = [1] * size

    def find(self, x):
        if x == self.root[x]:
            return x
        self.root[x] = self.find(self.root[x])
        return self.root[x]

    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)
        if rootX != rootY:
            if self.rank[rootX] > self.rank[rootY]:
                self.root[rootY] = rootX
            elif self.rank[rootX] < self.rank[rootY]:
                self.root[rootX] = rootY
            else:
                self.root[rootY] = rootX
                self.rank[rootX] += 1

    def connected(self, x, y):
        return self.find(x) == self.find(y)
    


class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        if not points or len(points) == 0:
            return 0
        
        size = len(points)
        pq = []
        uf = UnionFind(size)

        for i in range(size):
            x1, y1 = points[i]
            for j in range(i + 1, size):
                x2, y2 = points[j]
                cost = abs(x1 - x2) + abs(y1 - y2)
                edge = Edge(i, j, cost)
                pq.append(edge)
        
        heapq.heapify(pq)

        result = 0
        count = size - 1
        while pq and count > 0:
            edge = heapq.heappop(pq)
            if not uf.connected(edge.point1, edge.point2):
                uf.union(edge.point1, edge.point2)
                result += edge.cost
                count -= 1
        return result
                    
                    
        
108351