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Author: Chris Wu

problems/accounts-merge.py

@@ -0,0 +1,52 @@
+from collections import defaultdict
+
+class Solution(object):
+    def accountsMerge(self, accounts):
+        graph = defaultdict(list)
+        merged = set()
+        ans = []
+
+        #[0]
+        for data in accounts:
+            emails = data[1:]
+            for i, email in enumerate(emails):
+                graph[email].extend(emails[:i])
+                graph[email].extend(emails[i+1:])
+        
+        for data in accounts:
+            name = data[0]
+            visited = set()
+            stack = [data[1]]
+
+            if data[1] in merged: continue #[1]
+            
+            while stack:
+                e = stack.pop()
+                if e in visited: continue
+                visited.add(e)
+                stack.extend(graph[e])
+            
+            merged.update(visited)
+            ans.append([name]+sorted(list(visited)))
+        
+        return ans
+
+"""
+Treat each email as a node.
+Build an adjacency graph. [0]
+
+For every account's data
+First, lets check if the first email is already merged. [1]
+If the first email is already merged to other groups, then other emails will be in another group as well.
+So don't need to check.
+
+Second, do a DFS starting from the first email. Put all the connected nodes into visited.
+And append it to the ans with name.
+
+Let N be the total number of emails. M be the total number of final groups.
+Build the graph takes O(N).
+For each final groups, we make a DFS to all nodes, taking O(MN).
+Sort the group takes O((N/M)*Log(N/M)) -> O((N/M)*(logN-LogM))
+Time: O(MN)
+Space: O(N)
+"""