<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){
        // we have taken int find because, here we have to return the value of the par
        if(x==parent[x]){
            return x;
            // if A ka parent A hi hai, toh A ko hi return kar lena
        }
            return parent[x]=find(parent[x]);
            //if A ka parent A ke alava kuchaur h toh maan lo ki A ka parent B h,
            //so B ka parent find karlo adn return karlo

    }
    void Union(int x,int y){//Union ka U captital honi chahiye. Kyu ki small union
    //yek inbuilt function h c++ ka
        int x_parent = find(x);
        int y_parent = find(y);
        if(x_parent == y_parent){
            return;
        }
        if(rank[x_parent]>rank[y_parent]){
            //if x_parent ka rank y_parent ka rank se bhi bada hai, maane ki x hi 
            //x_parent hi bada h, so usko hi y_parent ka parent ban lo.
            parent[y_parent] = x_parent;

        }
        else if (rank[x_parent]<rank[y_parent]){
            parent[x_parent] = y_parent;
        }
        else{
            //if dono ka rank equal ho to kisi yek ko parent ban lo. mei y ko bana
            //lethi hu.....
            parent[x_parent]=y_parent;
            rank[y_parent]++;   
        }
    }
    bool equationsPossible(vector<string>& equations) {
        parent.resize(26);
        rank.resize(26,0);
        for(int i =0;i<26;i++){
            parent[i]=i; // starting mei khud ka parent khud hi honge
        }
        for(string & s:equations){
            if(s[1]=='='){//a==b
                Union(s[0]-'a',s[3]-'a');
            }
        }
        for(string & s:equations){
            if(s[1]=='!'){//a!=b
                char first = s[0];
                char second = s[3];
                int parent_first = find(first - 'a');
                int parent_second = find(second - 'a');
                if(parent_first==parent_second){
                    return false;
                }
                
            }
        }
        return true;

    }
};
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