<p>You are given an array of strings <code>equations</code> that represent relationships between variables where each string <code>equations[i]</code> is of length <code>4</code> and takes one of two different forms: <code>&quot;x<sub>i</sub>==y<sub>i</sub>&quot;</code> or <code>&quot;x<sub>i</sub>!=y<sub>i</sub>&quot;</code>.Here, <code>x<sub>i</sub></code> and <code>y<sub>i</sub></code> are lowercase letters (not necessarily different) that represent one-letter variable names.</p>

<p>Return <code>true</code><em> if it is possible to assign integers to variable names so as to satisfy all the given equations, or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> equations = [&quot;a==b&quot;,&quot;b!=a&quot;]
<strong>Output:</strong> false
<strong>Explanation:</strong> If we assign say, a = 1 and b = 1, then the first equation is satisfied, but not the second.
There is no way to assign the variables to satisfy both equations.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> equations = [&quot;b==a&quot;,&quot;a==b&quot;]
<strong>Output:</strong> true
<strong>Explanation:</strong> We could assign a = 1 and b = 1 to satisfy both equations.
</pre>

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

<ul>
	<li><code>1 &lt;= equations.length &lt;= 500</code></li>
	<li><code>equations[i].length == 4</code></li>
	<li><code>equations[i][0]</code> is a lowercase letter.</li>
	<li><code>equations[i][1]</code> is either <code>&#39;=&#39;</code> or <code>&#39;!&#39;</code>.</li>
	<li><code>equations[i][2]</code> is <code>&#39;=&#39;</code>.</li>
	<li><code>equations[i][3]</code> is a lowercase letter.</li>
</ul>

class UF:
    def __init__(self, n):
        self.parents = list(range(n))
        self.ranks = [1] * n
    def find(self, a):
        if a != self.parents[a]:
            self.parents[a] = self.find(self.parents[a])
        return self.parents[a]
    def union(self, a, b):
        p1, p2 = self.find(a), self.find(b)
        if p1 != p2:
            if self.ranks[p1] > self.ranks[p2]:
                self.parents[p2] = p1
                self.ranks[p1] += self.ranks[p2]
            else:
                self.parents[p1] = p2
                self.ranks[p2] += self.ranks[p1]
        
class Solution:
    def equationsPossible(self, equations: List[str]) -> bool:
        letters = {}
        eqs = []
        non = []
        for eq in equations:
            letters.setdefault(eq[0], len(letters))
            letters.setdefault(eq[3], len(letters))
            if eq[1] == "=":
                eqs.append([letters[eq[0]], letters[eq[3]]])
            else:
                non.append([letters[eq[0]], letters[eq[3]]])
        uf = UF(len(letters))

        for eq in eqs:
            uf.union(eq[0], eq[1])
        
        for n in non:
            if uf.find(n[0]) == uf.find(n[1]):
                return False
        return True
        
16800