<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, pa: List[List[int]]) -> str:
        p=[0]*(len(s))
        r=[0]*(len(s))
        for i in range (len(p)):
            p[i]=i
        def find(x):
            if p[x]==x:
                return x
            return find(p[x])
        def union(x,y):
            p1,p2=find(x),find(y)
            if p1!=p2:
                if r[p1]>r[p2]:
                    p[p2]=p1
                elif r[p1]<r[p2]:
                    p[p1]=p2
                else:
                    p[p1]=p2
                    r[p2]+=1
        for i in pa:
            union(i[0],i[1])
        a={}
        for i in range (len(p)):
            if a.get(find(i)):
                a[find(i)].append(s[i])
            else:
                a[find(i)]=[]
                a[find(i)].append(s[i])
        for i in a:
            a[i].sort()
        s=""
        for i in range (len(p)):
            s+=a[find(i)][0]
            a[find(i)].pop(0)
        return s


        
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