<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 Solution {
public:
    vector<int> parent;
    vector<int> rank;
    int find(int x){
        if(parent[x] == x)
            return parent[x];
        
        return parent[x] = find(parent[x]);
    }

    void Union(int x,int y){
        int parent_x = find(x);
        int parent_y = find(y);

        if(rank[parent_x] > rank[parent_y]){
            parent[parent_y] = parent_x;
        }else if(rank[parent_y] > rank[parent_x]){
            parent[parent_x] = parent_y;
        }else{
            parent[parent_x] = parent_y;
            rank[parent_y]++;
        }
    }
    bool equationsPossible(vector<string>& equations) {
        parent.resize(26);
        rank.resize(26, 1);

        for(int i = 0; i < 26; i++){
            parent[i] = i;
        }

        for(string &s : equations){
            if(s[1] == '='){
                Union(s[0] - 'a', s[3] - 'a');
            }
        }

        for(string &s : equations){
            if(s[1] == '!'){
                if(find(s[0]-'a') == find(s[3] - 'a'))
                    return false;
            }
        }

        return true;
    }
};
15000