<p>You are given a string <code>s</code>, and an array of pairs of indices in the string&nbsp;<code>pairs</code>&nbsp;where&nbsp;<code>pairs[i] =&nbsp;[a, b]</code>&nbsp;indicates 2 indices(0-indexed) of the string.</p>

<p>You can&nbsp;swap the characters at any pair of indices in the given&nbsp;<code>pairs</code>&nbsp;<strong>any number of times</strong>.</p>

<p>Return the&nbsp;lexicographically smallest string that <code>s</code>&nbsp;can be changed to after using the swaps.</p>

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

<pre>
<strong>Input:</strong> s = &quot;dcab&quot;, pairs = [[0,3],[1,2]]
<strong>Output:</strong> &quot;bacd&quot;
<strong>Explaination:</strong> 
Swap s[0] and s[3], s = &quot;bcad&quot;
Swap s[1] and s[2], s = &quot;bacd&quot;
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;dcab&quot;, pairs = [[0,3],[1,2],[0,2]]
<strong>Output:</strong> &quot;abcd&quot;
<strong>Explaination: </strong>
Swap s[0] and s[3], s = &quot;bcad&quot;
Swap s[0] and s[2], s = &quot;acbd&quot;
Swap s[1] and s[2], s = &quot;abcd&quot;</pre>

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

<pre>
<strong>Input:</strong> s = &quot;cba&quot;, pairs = [[0,1],[1,2]]
<strong>Output:</strong> &quot;abc&quot;
<strong>Explaination: </strong>
Swap s[0] and s[1], s = &quot;bca&quot;
Swap s[1] and s[2], s = &quot;bac&quot;
Swap s[0] and s[1], s = &quot;abc&quot;
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10^5</code></li>
	<li><code>0 &lt;= pairs.length &lt;= 10^5</code></li>
	<li><code>0 &lt;= pairs[i][0], pairs[i][1] &lt;&nbsp;s.length</code></li>
	<li><code>s</code>&nbsp;only contains lower case English letters.</li>
</ul>

class Solution {
public:
    
       int find_uparent(int x, vector<int>& parent)
       {
        if(parent[x]==x) return x;
        return parent[x]=find_uparent(parent[x], parent);
    }
    string smallestStringWithSwaps(string s, vector<vector<int>>& pairs) {
        // find all elements that belong to ne group

        //union find implementation
        int str_len = s.size();
        vector<int> parent(str_len);
        vector<int> rank(str_len, 0);

        for(int i=0; i<str_len; i++)
        {
            parent[i] = i;
        }

        for(int i=0; i<pairs.size(); i++)
        {
            int first = pairs[i][0];
            int second = pairs[i][1];

            int p1 = find_uparent(first, parent);
            int p2 = find_uparent(second, parent);
            // return "";
            if(p1 != p2)
            {
                if(rank[p1] >= rank[p2])
                {
                    parent[p2] = p1;
                    rank[p1]+=rank[p2];
                    // cout<<" loop 1 "<<parent[p2]<<" "<<p2;
                }
                else
                {
                    parent[p1] =p2;
                    rank[p2]+=rank[p1];
                    cout<<" loop 2"<<parent[p1]<<" "<<p2;
                }

            }
            
        }
        // return "";
                //doing path compression, while doing a find
        
        for(int i=0; i<str_len; i++)
        {
           parent[i] = find_uparent(i, parent);
           cout<<"i and parent is: "<<i<<" "<<parent[i]<<" ";
        }


        // map<int, vector<char>> m1;
        // for(int i=0; i<str_len; i++)
        // {
        //    m1[parent[i]].push_back(s[i]);
        // }

        // //sort the vector in map
        // string final="";
        // for(auto it: m1)
        // {
        //     vector<char> temp = it.second;
        //     sort(temp.begin(),temp.end());
        //     for(int k=0; k<temp.size(); k++) final+=temp[k];
        //     // m1[it.first] = temp;
        // }

        // return final;

        //we have sorted vectors of the size of map, we need to merge them, merge based off the first element

        //we have to do map and sorting, so will use a priority queue

        unordered_map<int, priority_queue<int, vector<int>, greater<int>>> m1;
        unordered_map<int, priority_queue<char, vector<char>, greater<char>>> m2;

        for(int i=0; i<str_len; i++)
        {
            m1[parent[i]].push(i);
            m2[parent[i]].push(s[i]);
        }

        //within connected components sorting done now reassiging the indices

        int i=0;
        while(s[i] != '\0')
        {
            if(m1.find(i) != m1.end())
            {
                while(m1[i].size())
                {
                    int top_ind = m1[i].top();
                    char top_val = m2[i].top();
                    m1[i].pop();
                    m2[i].pop();
                    cout<<top_ind<<" "<<top_val<<" ";
                    s[top_ind] = top_val;
                }
            }
            i++;
        }

        return s;
    }
};
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