Sync LeetCode submission Runtime - 30 ms (52.74%), Memory - 26.1 MB (8.35%)

Author: jimit105

0721-accounts-merge/README.md

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+<p>Given a list of <code>accounts</code> where each element <code>accounts[i]</code> is a list of strings, where the first element <code>accounts[i][0]</code> is a name, and the rest of the elements are <strong>emails</strong> representing emails of the account.</p>
+
+<p>Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some common email to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.</p>
+
+<p>After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails <strong>in sorted order</strong>. The accounts themselves can be returned in <strong>any order</strong>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> accounts = [[&quot;John&quot;,&quot;johnsmith@mail.com&quot;,&quot;john_newyork@mail.com&quot;],[&quot;John&quot;,&quot;johnsmith@mail.com&quot;,&quot;john00@mail.com&quot;],[&quot;Mary&quot;,&quot;mary@mail.com&quot;],[&quot;John&quot;,&quot;johnnybravo@mail.com&quot;]]
+<strong>Output:</strong> [[&quot;John&quot;,&quot;john00@mail.com&quot;,&quot;john_newyork@mail.com&quot;,&quot;johnsmith@mail.com&quot;],[&quot;Mary&quot;,&quot;mary@mail.com&quot;],[&quot;John&quot;,&quot;johnnybravo@mail.com&quot;]]
+<strong>Explanation:</strong>
+The first and second John&#39;s are the same person as they have the common email &quot;johnsmith@mail.com&quot;.
+The third John and Mary are different people as none of their email addresses are used by other accounts.
+We could return these lists in any order, for example the answer [[&#39;Mary&#39;, &#39;mary@mail.com&#39;], [&#39;John&#39;, &#39;johnnybravo@mail.com&#39;], 
+[&#39;John&#39;, &#39;john00@mail.com&#39;, &#39;john_newyork@mail.com&#39;, &#39;johnsmith@mail.com&#39;]] would still be accepted.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> accounts = [[&quot;Gabe&quot;,&quot;Gabe0@m.co&quot;,&quot;Gabe3@m.co&quot;,&quot;Gabe1@m.co&quot;],[&quot;Kevin&quot;,&quot;Kevin3@m.co&quot;,&quot;Kevin5@m.co&quot;,&quot;Kevin0@m.co&quot;],[&quot;Ethan&quot;,&quot;Ethan5@m.co&quot;,&quot;Ethan4@m.co&quot;,&quot;Ethan0@m.co&quot;],[&quot;Hanzo&quot;,&quot;Hanzo3@m.co&quot;,&quot;Hanzo1@m.co&quot;,&quot;Hanzo0@m.co&quot;],[&quot;Fern&quot;,&quot;Fern5@m.co&quot;,&quot;Fern1@m.co&quot;,&quot;Fern0@m.co&quot;]]
+<strong>Output:</strong> [[&quot;Ethan&quot;,&quot;Ethan0@m.co&quot;,&quot;Ethan4@m.co&quot;,&quot;Ethan5@m.co&quot;],[&quot;Gabe&quot;,&quot;Gabe0@m.co&quot;,&quot;Gabe1@m.co&quot;,&quot;Gabe3@m.co&quot;],[&quot;Hanzo&quot;,&quot;Hanzo0@m.co&quot;,&quot;Hanzo1@m.co&quot;,&quot;Hanzo3@m.co&quot;],[&quot;Kevin&quot;,&quot;Kevin0@m.co&quot;,&quot;Kevin3@m.co&quot;,&quot;Kevin5@m.co&quot;],[&quot;Fern&quot;,&quot;Fern0@m.co&quot;,&quot;Fern1@m.co&quot;,&quot;Fern5@m.co&quot;]]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= accounts.length &lt;= 1000</code></li>
+	<li><code>2 &lt;= accounts[i].length &lt;= 10</code></li>
+	<li><code>1 &lt;= accounts[i][j].length &lt;= 30</code></li>
+	<li><code>accounts[i][0]</code> consists of English letters.</li>
+	<li><code>accounts[i][j] (for j &gt; 0)</code> is a valid email.</li>
+</ul>

0721-accounts-merge/solution.py

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+# Approach 1: Depth First Search
+
+# n = no. of accounts, k = max length of account
+# Time: O(n * k * log (nk))
+# Space: O(nk)
+
+class Solution:
+    def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
+        # create email to name mapping
+        email_to_name = {}
+        # create email graph
+        email_graph = {}
+
+        # Build the graph
+        for account in accounts:
+            name = account[0]
+            # process all emails in the account
+            for email in account[1:]:
+                email_to_name[email] = name
+                if email not in email_graph:
+                    email_graph[email] = set()
+
+                # connect first email with all other emails in the account
+                if len(account) > 2:
+                    email_graph[account[1]].add(email)
+                    email_graph[email].add(account[1])
+
+        # DFS function to find all connected emails
+        def dfs(email, visited, connected_emails):
+            visited.add(email)
+            connected_emails.append(email)
+            for neighbor in email_graph[email]:
+                if neighbor not in visited:
+                    dfs(neighbor, visited, connected_emails)
+
+        # Process all emails and merge accounts
+        visited = set()
+        merged_accounts = []
+
+        for email in email_graph:
+            if email not in visited:
+                connected_emails = []
+                dfs(email, visited, connected_emails)
+                merged_accounts.append([email_to_name[email]] + sorted(connected_emails))
+
+        return merged_accounts