<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 heapq import heappush, heappop

class DisjointSet:
    def __init__(self, size):
        self.root = [0] * size
        for i in range(size):
            self.root[i] = i
        self.size = size
        
    def find(self, x):
        if self.root[x] == x:
            return x
        self.root[x] = self.find(self.root[x])
        return self.root[x]
    
    def union(self, x, y) -> bool:
        rx = self.find(x)
        ry = self.find(y)
        
        if rx == ry:
            return False
        
        self.root[rx] = ry
        return True
        
    

class Solution:
    def manhattanDistance(self, x1, y1, x2, y2) -> int:
        return abs(x2-x1) + abs(y2-y1)
    
    def minCostConnectPoints(self, p: List[List[int]]) -> int:
        ds = DisjointSet(len(p))
        heap = []
        
        for i, (x,y) in enumerate(p):
            for j, (x1, y1) in enumerate(p):
                if i != j:
                    heappush(heap, (self.manhattanDistance(x, y, x1, y1), i, j))
                
        total = 0
        
        while heap:
            dist, i, j = heappop(heap)
            
            if ds.union(i, j):
                total += dist
        
        return total
                
        
        
        
141030