<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:

string findp(string s,unordered_map<string,string>& parent){
    if(parent[s]==s) return s;
    return parent[s]=findp(parent[s],parent);
}
void unionf(string a,string b,unordered_map<string,string>& parent,unordered_map<string ,int>& rank){

   string pa=findp(a,parent);
   string pb=findp(b,parent);
   if(pa==pb) return ;
   if(rank[pa]<rank[pb]){
    parent[pa]=pb;
   }
   else if(rank[pa]>rank[pb]){
    parent[pb]=pa;
   }
   else{
    parent[pa]=pb;
    rank[pb]++;
   }
}
    bool equationsPossible(vector<string>& equations) {
        unordered_map<string,string> parent;
        unordered_map<string ,int> rank;
        for(auto& it:equations){
            string x=string(1,it[0]);
            string y=string(1,it[3]);
            if(parent.find(x)==parent.end()){
                parent[x]=x;
                rank[x]=0;
            }
             if(parent.find(y)==parent.end()){
                parent[y]=y;
                rank[y]=0;
            }

        }
        for(auto &it:equations){
            if(it[1]=='='){
                if(findp(string(1,it[0]),parent)==findp(string(1,it[3]),parent))continue;
                unionf(string(1,it[0]),string(1,it[3]),parent,rank);

            
            }
        }
        for(auto &it:equations){
            if(it[1]=='!'){
                if(findp(string(1,it[0]),parent)==findp(string(1,it[3]),parent)) return false;
            }
        }
        return true;
    }
};
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