<p>There are <code>n</code> computers numbered from <code>0</code> to <code>n - 1</code> connected by ethernet cables <code>connections</code> forming a network where <code>connections[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> represents a connection between computers <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>. Any computer can reach any other computer directly or indirectly through the network.</p>

<p>You are given an initial computer network <code>connections</code>. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected.</p>

<p>Return <em>the minimum number of times you need to do this in order to make all the computers connected</em>. If it is not possible, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/01/02/sample_1_1677.png" style="width: 500px; height: 148px;" />
<pre>
<strong>Input:</strong> n = 4, connections = [[0,1],[0,2],[1,2]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> Remove cable between computer 1 and 2 and place between computers 1 and 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/01/02/sample_2_1677.png" style="width: 500px; height: 129px;" />
<pre>
<strong>Input:</strong> n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 6, connections = [[0,1],[0,2],[0,3],[1,2]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There are not enough cables.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= connections.length &lt;= min(n * (n - 1) / 2, 10<sup>5</sup>)</code></li>
	<li><code>connections[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>There are no repeated connections.</li>
	<li>No two computers are connected by more than one cable.</li>
</ul>

class UF:
    def __init__(self, n):
        self.par = {}
        self.rank = {}

        for i in range(n):
            self.par[i] = i
            self.rank[i] = 1
        
    def find(self, n):
        while self.par[n] != n:
            self.par[n] = self.par[self.par[n]]
            n = self.par[n]
        return n
    
    def union(self, n1, n2):
        p1, p2 = self.find(n1), self.find(n2)
        if p1 == p2: return False

        if self.rank[p1] == self.rank[p2]: 
            self.par[p2] = p1
            self.rank[p1] += 1
        elif self.rank[p1] > self.rank[p2]:
            self.par[p2] = p1
        else:
            self.par[p1] = p2

        return True

class Solution:
    def makeConnected(self, n: int, connections: List[List[int]]) -> int:
        extra_cables = 0
        uf = UF(n)
        srvrs = set()

        for n1, n2 in connections:
            srvrs.add(n1)
            srvrs.add(n2)
            if uf.union(n1, n2) == False: extra_cables += 1
            else: n -= 1

        if extra_cables >= n-1:
            return n - 1
        return -1
37600