<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:
    map<int,multiset<char>>mp;

    int findp(int i,vector<int>&par){
        if(par[i]==i){
            return i;
        }
        return par[i]=findp(par[i],par);
    }
    void unionf(int a,int b,vector<int>&rank,vector<int>&par){

        a=findp(a,par);
        b=findp(b,par);
        //dsu union i s of 2 types 
        //union  by rank 
        //and union by value
        //here we will be using union by rank

        if(a!=b){

            if(rank[a]<rank[b]){
                par[a]=b;


            }else if(rank[a]>rank[b]){
                par[b]=a;

            }else{
                par[a]=b;
                rank[b]++;
            }
        }
    }
    string smallestStringWithSwaps(string s, vector<vector<int>>& pairs) {



        //how the inution will go about

        int n=s.size();
        vector<int>rank(n,1);
        vector<int>par(n,0);
        for(int i=0;i<n;i++){
            rank[i]=1;
            par[i]=i;
        }

        for(auto e:pairs){
            int a=e[0];
            int b=e[1];
            unionf(a,b,rank,par);
        }

        //har index ke char ko apne  connected componenet ke ultimate parent ware multiset me daal donga
        for(int i=0;i<s.size();i++){
            int p=findp(i,par);
            mp[p].insert(s[i]);
        }
        //here we will be applying union find on indexs which are connected ie 
        //which can be swapped will be part of the same connected component


        //now i will try making the lexicographically smallest substring 
        //from the data struture that i have made
        string ans="";
        for(int i=0;i<s.size();i++){
            int p=findp(i,par);
            char c=*mp[p].begin();
            mp[p].erase(mp[p].begin());
            ans+=c;
        }
        return ans;





        //in this question we will be using the concept of dsu along wiht priority queue 



        
    }
};
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