<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:

    //DFS appraoch:  
    void dfs(char u, unordered_map<char, vector<char>>& adj, vector<bool>& vis, char& minChar){
      vis[u-'a']=true;
      minChar=min(minChar, u);
      for(char v: adj[u]){
        if(!vis[v-'a']){
          dfs(v, adj, vis, minChar);
        }
      }
}

    string smallestEquivalentString(string s1, string s2, string baseStr) {
        
      int n=s1.size();
      unordered_map<char, vector<char>> adj;
      for(int i=0; i<n; i++){
        char u=s1[i];
        char v=s2[i];
        adj[u].push_back(v);
        adj[v].push_back(u);
      }

      int m=baseStr.size();
      string ans="";
      for(int i=0; i<m; i++){
        vector<bool> vis(26, false);
        char ch=baseStr[i];
        char minChar=ch;
        dfs(ch, adj, vis, minChar);
        ans+=minChar;
      }

      return ans;
    }
};



//Using BFS approach:
/*
class Solution{
public:

    //BFS logic:
    char bfs(char u, unordered_map<char, vector<char>>& adj){
      
      char minChar=u;
      vector<bool> vis(26, false);
      queue<char> q;
      q.push(u);
      vis[u-'a']=true;
      while(!q.empty()){
        char node=q.front();
        q.pop();
        minChar=min(minChar, node);
        for(char v: adj[node]){
          if(!vis[v-'a']){
            vis[v-'a']=true;
            q.push(v);
          }
        }
      }
      return minChar;
}

    //Question function:
    string smallestEquivalentString(string s1, string s2, string baseStr) {
        
      int n=s1.size();
      unordered_map<char, vector<char>> adj;
      for(int i=0; i<n; i++){
        char u=s1[i];
        char v=s2[i];
        adj[u].push_back(v);
        adj[v].push_back(u);
      }

      int m=baseStr.size();
      string ans="";
      for(int i=0; i<m; i++){
        vector<bool> vis(26, false);
        char ch=baseStr[i];
        char minChar=bfs(ch, adj);
        ans+=minChar;
      }

      return ans;
    }
};
*/
11500