<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:
        def findUnion(i,n,visited,d1):
            ans = set()
            q = [i]
            while(q):
                curr = q[0]
                q.pop(0)
                ans.add(curr)
                visited[curr] = True
                for j in d1[curr]:
                    if(visited[j]==False):
                        q.append(j)
            return ans
        if(not pairs):
            return s
        s = list(s)
        n = max( max([pair[0] for pair in pairs]), max([pair[1] for pair in pairs]))
        visited = [False for i in range(n+1)]
        d1 = defaultdict(list)
        for pair in pairs:
            a = pair[0]
            b = pair[1]
            d1[a].append(b)
            d1[b].append(a)
        for i in range(n):
            if(visited[i]==False):
                union = list(findUnion(i,n,visited,d1))
                union.sort()
                union_chr = [s[j] for j in union]
                union_chr.sort()
                k = 0
                for j in union:

                    s[j] = union_chr[k]
                    k+=1
        return "".join(s)
816