<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, size):
        self.parent = defaultdict(lambda: -1)
        self.rank = defaultdict(int)
        self.size = size

    def find(self, x):
        if self.parent[x] == -1:
            return x
        self.parent[x] = self.find(self.parent[x]) 
        return self.parent[x]

    def union(self, x, y): 
        parent1 = self.find(x) 
        parent2 = self.find(y)
        if parent1 == parent2:
            return False
        if self.rank[parent1] < self.rank[parent2]:
            self.parent[parent1] = parent2
            self.rank[parent2] += 1
        else:
            self.parent[parent2] = parent1
            self.rank[parent1] += 1
        self.size -= 1
        return True 

class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        sorted_costs = []
        for i in range(len(points)):
            first_point = points[i]
            for j in range(i, len(points)):
                second_point = points[j]
                manhattan_dist = abs(first_point[0] - second_point[0]) + abs(first_point[1] - second_point[1])
                sorted_costs.append((manhattan_dist, first_point, second_point))

        sorted_costs.sort()

        connected_points = UnionFind(len(points))
        min_cost = 0 
        for dist, first, second in sorted_costs: 
            if connected_points.union(tuple(first), tuple(second)):
                min_cost += dist
        return min_cost
            
        
        

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