<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 UnionFind {
public:
    UnionFind(int sz) : root(sz), rank(sz), count(sz) {
        for (int i = 0; i < sz; i++) {
            root[i] = i;
            rank[i] = 1;
        }
    }

    int find(int x) {
        if (x == root[x]) {
            return x;
        }
	// Some ranks may become obsolete so they are not updated
        return root[x] = find(root[x]);
    }

    void unionSet(int x, int y) {
        int rootX = find(x);
        int rootY = find(y);
        if (rootX != rootY) {
            if (rank[rootX] > rank[rootY]) {
                root[rootY] = rootX;
            } else if (rank[rootX] < rank[rootY]) {
                root[rootX] = rootY;
            } else {
                root[rootY] = rootX;
                rank[rootX] += 1;
            }
            count--;
        }
    }

    bool connected(int x, int y) {
        return find(x) == find(y);
    }
    
    int getCount(void) {
        return count;
    }

private:
    vector<int> root;
    vector<int> rank;
    int count;
};

class Solution {
public:
    string smallestStringWithSwaps(string s, vector<vector<int>>& pairs) {
        int n = s.length();
        UnionFind uf(n);
        
        for (vector<int> &p : pairs) {
            uf.unionSet(p[0], p[1]);
        }
        
        unordered_map<int,string> m;
        
        for (int i = 0; i < n; i++) {
            int root = uf.find(i);
            m[root] += s[i];
        }
        
        if (uf.getCount() == 1) {
            sort(s.begin(), s.end());
            return s;
        }
        
        
        for (auto pairs : m) {
            int i = pairs.first;
            string s = pairs.second;
            
            sort(s.begin(), s.end(), greater<char>());
            
            m[i] = s;
        }
        
        for (int i = 0; i < n; i++) {
            int root = uf.find(i);
            s[i]=m[root].back();
            m[root].pop_back();
        }
        
        return s;
    }
};
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