<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 {
public:
    void dfs(vector<char> adj[], char node, char minCharacter, vector<char>& visited)
    {
        // to this character, replace it with it's minimum character
        visited[node - 'a'] = minCharacter;
        
        // travel in neighbour of this character,
        // i.e travel in component of this character
        for(char& near: adj[node - 'a'])
        {
            // if this character is not visited, call dfs for it
            if(visited[near - 'a'] == '#')
            {
                dfs(adj, near, minCharacter, visited);
            }
        }
    }
    
    string smallestEquivalentString(string s1, string s2, string baseStr) {
        int n = s1.size(); // extract size
        
        vector<char> adj[26]; // declaring adjacency matrix
        
        // travel through strings and establish connection
        for(int i = 0; i < n; i++)
        {
            adj[s1[i] - 'a'].push_back(s2[i]);
            adj[s2[i] - 'a'].push_back(s1[i]);
        }
        
        // make a visited vector of size 26
        // and symbol, '#' denotes that this particular character is not visited yet
        vector<char> visited(26, '#');
        
        // now, travel through each character
        for(char c = 'a'; c <= 'z'; c++)
        {
            // if this character is not visited yet, call dfs here 
            if(visited[c - 'a'] == '#')
            {
                // as we are calling dfs for each character, so
                // it will minimum value for that component.
                dfs(adj, c, c, visited);
            }
        }
        
        // Now lastly, replace each baseStr character with their 
        // minimum found out character in their component,
        for(int i = 0; i < baseStr.size(); i++)
        {
            baseStr[i] = visited[baseStr[i] - 'a'];
        }
        
        // return baseStr now, 
        return baseStr;
    }
};
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