<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 UnionFind:
    def __init__(self):
        self.size = [1]*26
        self.parent = [i for i in range(26)]
    def findParent(self,x):
        while x != self.parent[x]:
            x = self.findParent(self.parent[x])
        return x
    def union(self,x, y):
        parent_x = self.findParent(x)
        parent_y = self.findParent(y)
        if parent_x == parent_y:
            return False
        if self.size[parent_x] > self.size[parent_y]:
            self.size[parent_x] += self.size[parent_y]
            self.parent[parent_y] = parent_x
        else:
            self.size[parent_y] += self.size[parent_x]
            self.parent[parent_x] = parent_y
        return True

    
class Solution:
    def equationsPossible(self, equations: List[str]) -> bool:
        uf = UnionFind()
        for equation in equations:
            if equation[1] == '=':
                char1 = equation[0]
                char2 = equation[3]
                uf.union(ord(char1)-ord('a'), ord(char2)-ord('a'))
        
        for equation in equations:
            if equation[1] == '!':
                char1 = equation[0]
                char2 = equation[3]
                if uf.findParent(ord(char1)-ord('a')) == uf.findParent(ord(char2)-ord('a')):
                    return False
        return True
16500