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+You are given a string `s`, and an array of pairs of indices in the string `pairs` where `pairs[i] = [a, b]` indicates 2 indices(0-indexed) of the string.
+
+You can swap the characters at any pair of indices in the given `pairs` **any number of times**.
+
+Return the lexicographically smallest string that `s` can be changed to after using the swaps.
+
+**Example 1:**
+
+    Input: s = "dcab", pairs = [[0,3],[1,2]]
+    Output: "bacd"
+    Explaination: 
+    Swap s[0] and s[3], s = "bcad"
+    Swap s[1] and s[2], s = "bacd"
+    
+
+**Example 2:**
+
+    Input: s = "dcab", pairs = [[0,3],[1,2],[0,2]]
+    Output: "abcd"
+    Explaination: 
+    Swap s[0] and s[3], s = "bcad"
+    Swap s[0] and s[2], s = "acbd"
+    Swap s[1] and s[2], s = "abcd"
+
+**Example 3:**
+
+    Input: s = "cba", pairs = [[0,1],[1,2]]
+    Output: "abc"
+    Explaination: 
+    Swap s[0] and s[1], s = "bca"
+    Swap s[1] and s[2], s = "bac"
+    Swap s[0] and s[1], s = "abc"
+    
+
+**Constraints:**
+
+*   `1 <= s.length <= 10^5`
+*   `0 <= pairs.length <= 10^5`
+*   `0 <= pairs[i][0], pairs[i][1] < s.length`
+*   `s` only contains lower case English letters.
+
+https://leetcode.com/problems/smallest-string-with-swaps
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diff --git a/1308_smallest-string-with-swaps/python3/solution.py b/1308_smallest-string-with-swaps/python3/solution.py
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+from collections import defaultdict
+
+class Solution:
+    def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
+        '''
+        Key idea: if (a, b) indexes can be swapped and (b, c) indexes can
+        be swapped, then (a, c) can be swapped so infinite swap across (a, b, c).
+        Can be expanded to cover any other letters that are swappable.
+        '''
+        
+        # Make the adjacency list
+        swapmap = defaultdict(list)
+        
+        for i, j in pairs:
+            swapmap[i].append(j)
+            swapmap[j].append(i)
+            
+        
+        def dfs(s, vertex, chars, vertices):
+            chars.append(s[vertex])
+            vertices.append(vertex)
+            
+            visited.add(vertex)
+            
+            for neighbor in swapmap[vertex]:
+                if neighbor not in visited:
+                    dfs(s, neighbor, chars, vertices)
+        
+        
+        output = [''] * len(s)
+        visited = set()
+        
+        for vertex, _ in enumerate(s):
+            if vertex not in visited:
+                connected_chars = []
+                connected_vertices = []
+                
+                dfs(s, vertex, connected_chars, connected_vertices)
+                
+                # This might seem confusing but remember that if we find one character
+                # with the `vertex` connected (i.e swappable) with other characters, then
+                # they can all be swapped with each other infinite times. Our goal is to
+                # get the lexicographically smallest string (i.e most sorted) from doing
+                # these allowed swaps. Hence, we can go ahead and safely sort the chars
+                # obtained from DFS of vertex.
+                connected_chars.sort()
+                connected_vertices.sort()
+                
+                for i, connected_vertex in enumerate(connected_vertices):
+                    output[connected_vertex] = connected_chars[i]
+        
+        return ''.join(output)