<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 Solution:
    def smallestEquivalentString(self, s1, s2, baseStr):
        n, s, dict1 = len(s1), "", {}

        def find(x):
            if x not in dict1:
                return x 
            else:
                if x != dict1[x]:
                    dict1[x] = find(dict1[x])
                return dict1[x]

        def union(x,y):
            a, b = find(x), find(y)

            if a > b:
                dict1[a] = b 
            else:
                dict1[b] = a

        for i in range(n):
            union(s1[i],s2[i])

        for i in baseStr:
            s += find(i)

        return s 
16700