<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:
    string smallestStringWithSwaps(string s, vector<vector<int>>& pairs) {
        int n = s.size();
        vector<bool> visited(n, false); 
        unordered_map<int, vector<int>> adj;
        for(auto v : pairs){
            adj[v[0]].push_back(v[1]);
            adj[v[1]].push_back(v[0]);
        }
        for(int i = 0; i < n; i++){
            if(visited[i]) continue;
            // BFS
            queue<int> nodes;
            nodes.push(i);
            visited[i] = true;
            vector<int> idx;
            vector<char> C;
            while(!nodes.empty()){
                int k = nodes.front();
                idx.push_back(k);
                C.push_back(s[k]);
                nodes.pop();
                for(auto t : adj[k]){
                    if(!visited[t]){
                        nodes.push(t);
                        visited[t] = true;
                    }
                }
            }
            
            sort(idx.begin(), idx.end());
            sort(C.begin(), C.end());
        
            for(int j = 0; j < idx.size(); j++){
                s[idx[j]] = C[j];
            }
        }
        return s;
    }
};
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