smallest-string-with-swaps py3

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)