<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:
    int find(int i, vector<int>& v) {
        if (v[i] == i) {
            return i;
        }
        return v[i] = find(v[i], v);
    }

    void combine(char i, char j, vector<int> &v, vector<int> &size) {
        if ((i == j) || (v[i-'a'] == v[j-'a'])) {
            return;
        }

        int rootI = find(i-'a', v);
        int rootJ = find(j-'a', v);
        if (size[rootI] <= size[rootJ]) {
            v[rootI] = rootJ;
            size[rootJ] += size[rootI];
        } else {
            v[rootJ] = rootI;
            size[rootI] += size[rootI];
        }
    }

    bool equationsPossible(vector<string>& equations) {
        vector<int> v(27);
        vector<int> size(27);

        for (int i = 1; i <= 26; i++) {
            v[i] = i;
            size[i] = 1;
        }

        vector<string> equals;
        vector<string> notequals;

        for (auto equation : equations) {
            if (equation[1] == '!') {
                notequals.push_back(equation);
            } else {
                equals.push_back(equation);
            }
        }

        for (auto equation : equals) {
            char v1 = equation[0];
            char v2 = equation[3];
            combine(v1, v2, v, size);
        }

        for (auto equation : notequals) {
            int v1 = equation[0]-'a';
            char v2 = equation[3]-'a';
            int v1Root = find(v1, v);
            int v2Root = find(v2, v);

            cout<<v1Root<<" " << v2Root<<endl;
            if(v1Root == v2Root) {
                return false;
            }
        }

        return true;
    }
};
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