<p>You are given two strings of the same length <code>s1</code> and <code>s2</code> and a string <code>baseStr</code>.</p>

<p>We say <code>s1[i]</code> and <code>s2[i]</code> are equivalent characters.</p>

<ul>
	<li>For example, if <code>s1 = &quot;abc&quot;</code> and <code>s2 = &quot;cde&quot;</code>, then we have <code>&#39;a&#39; == &#39;c&#39;</code>, <code>&#39;b&#39; == &#39;d&#39;</code>, and <code>&#39;c&#39; == &#39;e&#39;</code>.</li>
</ul>

<p>Equivalent characters follow the usual rules of any equivalence relation:</p>

<ul>
	<li><strong>Reflexivity:</strong> <code>&#39;a&#39; == &#39;a&#39;</code>.</li>
	<li><strong>Symmetry:</strong> <code>&#39;a&#39; == &#39;b&#39;</code> implies <code>&#39;b&#39; == &#39;a&#39;</code>.</li>
	<li><strong>Transitivity:</strong> <code>&#39;a&#39; == &#39;b&#39;</code> and <code>&#39;b&#39; == &#39;c&#39;</code> implies <code>&#39;a&#39; == &#39;c&#39;</code>.</li>
</ul>

<p>For example, given the equivalency information from <code>s1 = &quot;abc&quot;</code> and <code>s2 = &quot;cde&quot;</code>, <code>&quot;acd&quot;</code> and <code>&quot;aab&quot;</code> are equivalent strings of <code>baseStr = &quot;eed&quot;</code>, and <code>&quot;aab&quot;</code> is the lexicographically smallest equivalent string of <code>baseStr</code>.</p>

<p>Return <em>the lexicographically smallest equivalent string of </em><code>baseStr</code><em> by using the equivalency information from </em><code>s1</code><em> and </em><code>s2</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s1 = &quot;parker&quot;, s2 = &quot;morris&quot;, baseStr = &quot;parser&quot;
<strong>Output:</strong> &quot;makkek&quot;
<strong>Explanation:</strong> Based on the equivalency information in s1 and s2, we can group their characters as [m,p], [a,o], [k,r,s], [e,i].
The characters in each group are equivalent and sorted in lexicographical order.
So the answer is &quot;makkek&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s1 = &quot;hello&quot;, s2 = &quot;world&quot;, baseStr = &quot;hold&quot;
<strong>Output:</strong> &quot;hdld&quot;
<strong>Explanation: </strong>Based on the equivalency information in s1 and s2, we can group their characters as [h,w], [d,e,o], [l,r].
So only the second letter &#39;o&#39; in baseStr is changed to &#39;d&#39;, the answer is &quot;hdld&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s1 = &quot;leetcode&quot;, s2 = &quot;programs&quot;, baseStr = &quot;sourcecode&quot;
<strong>Output:</strong> &quot;aauaaaaada&quot;
<strong>Explanation:</strong> We group the equivalent characters in s1 and s2 as [a,o,e,r,s,c], [l,p], [g,t] and [d,m], thus all letters in baseStr except &#39;u&#39; and &#39;d&#39; are transformed to &#39;a&#39;, the answer is &quot;aauaaaaada&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s1.length, s2.length, baseStr &lt;= 1000</code></li>
	<li><code>s1.length == s2.length</code></li>
	<li><code>s1</code>, <code>s2</code>, and <code>baseStr</code> consist of lowercase English letters.</li>
</ul>

class UnionFind {
private:
    int root[26] = { 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12,
                     13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25};
public:
    int Find(int node)
    {
        if(root[node] != node)
        {
            root[node] = Find(root[node]);
        }
        return root[node];
    }
    void Join(int n1, int n2)
    {
        int r1 = Find(n1);
        int r2 = Find(n2);
        if(r1 > r2)
        {
            root[r1] = r2;
        }
        else
        {
            root[r2] = r1;
        }
    }
};
class Solution {
public:
    string smallestEquivalentString(string s1, string s2, string baseStr)
    {
        UnionFind graph;
        for(int i = 0; i < s1.size(); i++)
        {
            graph.Join(s1[i] - 'a', s2[i] - 'a');
        }
        for(auto& ch: baseStr)
        {
            ch = 'a' + graph.Find(ch - 'a');
        }
        return baseStr;
    }
};

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