<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:
    # time = O(E + V log V)
    # space = O(V+E)
    def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
        graph = collections.defaultdict(list) # O(E) space
        for x,y in pairs: # O(E)
            graph[x].append(y)
            graph[y].append(x)
        
        visited = [False] * len(s) #O(V)
        def dfs(idx, chars, indices):
            if visited[idx]: return
            visited[idx] = True
            chars.append(s[idx])
            indices.append(idx)
            for neighbour in graph[idx]: # visit each connected edge O(E)
                dfs(neighbour, chars, indices)
        
        string = list(s) # as strings are not mutable in python
        for i in range(len(string)): # O(V log V) as each element is only sorted once (visited once)
            chars = []
            indices = []
            dfs(i, chars, indices)
            chars.sort() 
            indices.sort()
            
            for i in range(len(chars)):
                string[indices[i]] = chars[i] # smallest char assign to smallest index
        
        return "".join(string)


"""
dcab
0123

bcad
acbd
abcd

0 -> 2,3
1 -> 2
2 -> 0,1
3 -> 0


0 -- 3
  -- 2 --- 1

bcad
acbd


0 -- 1 -- 2
cba
bca
abc


edcba
eacbd
baced  

0-3
4-1
4-2

ebadc
01234

"""
582