<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:
#DFS
# we can swap any pair of vertices present in the same connected component.
        # s = [c for c in s]
        # n = len(s)
        # visited = set()
        # adj = defaultdict(list)
        # for u,v in pairs:
        #     adj[u].append(v)
        #     adj[v].append(u)

        # def dfs(node):
        #     chars.append(s[node])
        #     idxs.append(node)
        #     visited.add(node)
        #     for next_node in adj[node]:
        #         if next_node not in visited:
        #             dfs(next_node)

        # for i in range(n):
        #     if i not in visited:
        #         #chars and indexs in a connected component
        #         chars = []
        #         idxs = []
        #         dfs(i)
        #     # we can swap any pair of vertices present in the same connected component.
        #     # Thus, we simply sort chars and idxs to let the smallest idx have the smallest char
        #         chars.sort()
        #         idxs.sort()
        #         for i in range(len(idxs)):
        #             s[idxs[i]] = chars[i]
        # return ''.join(s)
        #Time:O(VlogV+E) Space:O(V+E)


#UnionFind
        s = [c for c in s]
        n = len(s)
        root = [i for i in range(n)]
        rank = [1]*n #Based on size
        def find(x):
            if x == root[x]:
                return x
            root[x] = find(root[x])
            return root[x]
        def union(x,y):
            rootX = find(x)
            rootY = find(y)
            if rootX != rootY:
                if rank[rootX] > rank[rootY]:
                    root[rootY] = rootX
                    rank[rootX] += rank[rootY]
                else:
                    root[rootX] = rootY
                    rank[rootY] += rank[rootX]
        for u,v in pairs:
            union(u,v)
        components = defaultdict(list)
    #Find all connected components list with the same root
        for i in range(n):
            root_i = find(i)
            components[root_i].append(i)

        for idx in components.values():
            chars = []
            for i in idx:
                chars.append(s[i])
            chars.sort()
            idx.sort()
            for i,c in zip(idx,chars):
                s[i] = c
        return ''.join(s)
        #Time:O((E+V)⋅α(V)+VlogV) Space:O(V)
471