<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 Dset:
#     def __init__(self, n):
#         self.root = [i for i in range(n)]
#         self.rank = [0] * n
    
#     def find(self, x: int) -> int:
#         if self.root[x] != x:
#             self.root[x] = self.find(self.root[x])
        
#         return self.root[x]
    
#     def union(self, x: int, y: int) -> None:
#         rootX = self.find(x)
#         rootY = self.find(y)

#         if rootX != rootY:
#             if self.rank[rootX] > self.rank[rootY]:
#                 self.root[rootY] = rootX
#             elif self.rank[rootY] < self.rank[rootX]:
#                 self.root[rootX] = rootY
#             else:
#                 self.root[rootY] = rootX
#                 self.rank[rootX] += 1


class Solution:
    def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
        
        # n = len(s)
        # components = defaultdict(list)
        # dset = Dset(n)

        # for a, b in pairs:
        #     dset.union(a, b)
        
        # for i in range(n):
        #     root = dset.find(i)
        #     components[root].append(s[i])
        
        # for key, value in components.items():
        #     value.sort(reverse = True)
        
        # newString = ''

        # for i in range(n):
        #     newString += components[dset.find(i)].pop()
        
        # return newString


        #dfs
        def dfs(i: int) -> list[int]:
            stack = [i]
            children = [i]
            
            while stack:
                node = stack.pop()
                for nNode in adj[node]:
                    if not visited[nNode]:
                        visited[nNode] = True
                        stack.append(nNode)
                        children.append(nNode)

            return children

        n = len(s)
        adj = defaultdict(set)
        components = list()
        visited = [False] * n 

        for a, b in pairs:
            adj[a].add(b)
            adj[b].add(a)
        
        for i in range(n):
            if not visited[i]:
                visited[i] = True
                components.append(dfs(i))
        
        result = list(s)
        for component in components:
            component.sort()
            chars = sorted(s[i] for i in component)

            for i, char in zip(component, chars):
                result[i] = char

        return ''.join(result)        



        
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