<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:
    class dsu{
    public:
        vector<int> par, rank;
        dsu() {
            rank.resize(26,1);
            for(int i=0; i<26; i++) {
                par.push_back(i);
            }
        }
        int find_(int x) {
            if(x == par[x]) {
                return x;
            }
            return par[x] = find_(par[x]);
        }
        int union_(int x, int y) {
            int px = find_(x), py = find_(y);
            if(px == py) {
                return 0;
            }
            if(rank[px] == rank[py]) {
                rank[px]++;
            }
            if(rank[px] > rank[py]) {
                par[py] = px;
                return 1;
            }
            par[px] = py;
            return 1;
        }
    };
    bool equationsPossible(vector<string>& eq) {
        dsu d;
        for(auto e: eq) {
            if(e[1] == '=')
                d.union_(e[0]-'a', e[3]-'a');
        }
        for(auto e: eq) {
            dsu d1 = d;
            if(e[1] == '!' && !d1.union_(e[0]-'a', e[3]-'a'))
                return false;
        }
        return true;
    }
};
17500