<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<vector<int>> graph(n);
        unordered_set<int> visited;
        vector<int> visited_idx;
        string cc;

        for (auto& pair : pairs) {
            graph[pair[0]].push_back(pair[1]);
            graph[pair[1]].push_back(pair[0]);
        }

        for (int i = 0; i < n; ++i) {
            if (visited.count(i) == 0) {
                cc.clear();
                visited_idx.clear();
                dfs(s, graph, cc, visited, visited_idx, i);
                sort(cc.begin(), cc.end());
                sort(visited_idx.begin(), visited_idx.end());
                for (int k = 0; k < cc.size(); ++k) {
                    s[visited_idx[k]] = cc[k];
                }
            }
        }

        return s;
    }
private:
    void dfs(string &s, vector<vector<int>> &graph, string &cc, unordered_set<int> &visited, vector<int> &visited_idx, int u) {
        visited.insert(u);
        visited_idx.push_back(u);
        cc += s[u];
        
        for (auto &v : graph[u]) {
            if (visited.count(v) == 0) {
                dfs(s, graph, cc, visited, visited_idx, v);
            }
        }

        return;
    }
};
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