<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>

# from scipy.sparse.csgraph import UnionFind

class UnionFind:
    def __init__(self, length):
        self.disjoint_sets = [-1] * length
    
    def find(self, index):
        # maybe do path compression
        if (self.disjoint_sets[index] == -1):
            return index
        
        root = self.find(self.disjoint_sets[index])
        self.disjoint_sets[index] = root
        return root

    def union(self, i1, i2):
        # union by rank?
        root_i1 = self.find(i1)
        root_i2 = self.find(i2)

        self.disjoint_sets[root_i1] = root_i2

    def connected(self, i1, i2):
        return self.find(i1) == self.find(i2)


class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        min_edges = 0 # (dist , from, to)
        edges = []
        cost = 0
        point_index = {} # point -> int (index)
        disjoint_set = UnionFind(len(points))

        for i in range(len(points)):
            point_index[tuple(points[i])] = i
            for j in range(i + 1, len(points)):
                x_diff = abs(points[i][0] - points[j][0])
                y_diff = abs(points[i][1] - points[j][1])
                edges.append((x_diff + y_diff, tuple(points[i]), tuple(points[j])))
        
        heapq.heapify(edges)
        
        while (min_edges != len(points) - 1):
            edge = heapq.heappop(edges) # (dis, from, to)
            set_1 = point_index[edge[1]]
            set_2 = point_index[edge[2]]

            if (not disjoint_set.connected(set_1, set_2)):
                disjoint_set.union(set_1, set_2)
                cost += edge[0]
                min_edges += 1
        
        return cost



        

        
130975