<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:
    def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
        """
        https://leetcode.com/problems/smallest-string-with-swaps/solutions/387524/short-python-union-find-solution-w-explanation/

        The core of the idea is that if (0, 1) is an exchange pair and (0, 2) is an exchange pair, then any 2 in (0, 1, 2) can be exchanged.
        """
        class UF:
            def __init__(self, n): 
                self.p = list(range(n))
            def union(self, x, y): 
                self.p[self.find(x)] = self.find(y)
            def find(self, x):
                if x != self.p[x]: 
                    self.p[x] = self.find(self.p[x])
                return self.p[x]
        uf, res, m = UF(len(s)), [], defaultdict(list)
        for x,y in pairs: 
            uf.union(x,y)
        for i in range(len(s)): 
            m[uf.find(i)].append(s[i])
        for comp_id in m.keys(): 
            m[comp_id].sort(reverse=True)
        for i in range(len(s)): 
            res.append(m[uf.find(i)].pop())
        return ''.join(res)
52805