<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:
    class UnionSet{
        public:
            vector<int> par;
            vector<int> siz;
            UnionSet(int n)
            {
                par.resize(n);
                siz.resize(n, 1);
                for(int i=0; i<n; i++)
                    par[i]=i;
            }

            int upar(int a)
            {
                if(a==par[a])
                    return a;
                
                return par[a]=upar(par[a]);
            }

            void join(int a, int b)
            {
                int upa=upar(a);
                int upb=upar(b);

                if(upa==upb)
                    return;
                if(siz[upa]>=siz[upb])
                {
                    siz[upa]+=siz[upb];
                    par[upb]=upa;
                }
                else
                {
                    siz[upb]+=siz[upa];
                    par[upa]=upb;
                }
            }
        unordered_map<char, vector<char>> grp()
        {
            unordered_map<char, vector<char>> mp;
            for(int i=0; i<26; i++)
                mp['a'+upar(i)].push_back('a'+i);
    
            for(auto i: mp)
                sort(mp[i.first].begin(), mp[i.first].end());

            return mp;
        }
    };
    string smallestEquivalentString(string s1, string s2, string baseStr) {
        
        UnionSet us(26);
        
        for(int i=0; i<s1.length(); i++)
            us.join(s1[i]-'a', s2[i]-'a');

        unordered_map<char, vector<char>> mp=us.grp();

        for(auto i: mp)
        {
            cout<<i.first<<"->";
            for(auto j: i.second)
                cout<<j<<" ";
            cout<<endl;
        }

        for(int i=0; i<baseStr.size(); i++)
        {
            char p=us.upar(baseStr[i]-'a')+'a';

            baseStr[i]=mp[p][0];
        }

        return baseStr;
        
    }
};
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