Sync LeetCode submission Runtime - 296 ms (49.26%), Memory - 18.7 MB (77.36%)

Author: jimit105

0127-word-ladder/README.md

@@ -0,0 +1,39 @@
+<p>A <strong>transformation sequence</strong> from word <code>beginWord</code> to word <code>endWord</code> using a dictionary <code>wordList</code> is a sequence of words <code>beginWord -&gt; s<sub>1</sub> -&gt; s<sub>2</sub> -&gt; ... -&gt; s<sub>k</sub></code> such that:</p>
+
+<ul>
+	<li>Every adjacent pair of words differs by a single letter.</li>
+	<li>Every <code>s<sub>i</sub></code> for <code>1 &lt;= i &lt;= k</code> is in <code>wordList</code>. Note that <code>beginWord</code> does not need to be in <code>wordList</code>.</li>
+	<li><code>s<sub>k</sub> == endWord</code></li>
+</ul>
+
+<p>Given two words, <code>beginWord</code> and <code>endWord</code>, and a dictionary <code>wordList</code>, return <em>the <strong>number of words</strong> in the <strong>shortest transformation sequence</strong> from</em> <code>beginWord</code> <em>to</em> <code>endWord</code><em>, or </em><code>0</code><em> if no such sequence exists.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> beginWord = &quot;hit&quot;, endWord = &quot;cog&quot;, wordList = [&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;lot&quot;,&quot;log&quot;,&quot;cog&quot;]
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> One shortest transformation sequence is &quot;hit&quot; -&gt; &quot;hot&quot; -&gt; &quot;dot&quot; -&gt; &quot;dog&quot; -&gt; cog&quot;, which is 5 words long.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> beginWord = &quot;hit&quot;, endWord = &quot;cog&quot;, wordList = [&quot;hot&quot;,&quot;dot&quot;,&quot;dog&quot;,&quot;lot&quot;,&quot;log&quot;]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> The endWord &quot;cog&quot; is not in wordList, therefore there is no valid transformation sequence.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= beginWord.length &lt;= 10</code></li>
+	<li><code>endWord.length == beginWord.length</code></li>
+	<li><code>1 &lt;= wordList.length &lt;= 5000</code></li>
+	<li><code>wordList[i].length == beginWord.length</code></li>
+	<li><code>beginWord</code>, <code>endWord</code>, and <code>wordList[i]</code> consist of lowercase English letters.</li>
+	<li><code>beginWord != endWord</code></li>
+	<li>All the words in <code>wordList</code> are <strong>unique</strong>.</li>
+</ul>

0127-word-ladder/solution.py

@@ -0,0 +1,35 @@
+# Approach: Breadth First Search
+
+# n = len(wordList), l = length of each word
+# Time: O(n * 26 * l)
+# Space: O(n)
+
+import string
+from collections import deque, defaultdict
+
+class Solution:
+    def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
+        if endWord not in wordList:
+            return 0
+
+        wordList = set(wordList)
+
+        queue = deque([(beginWord, 1)])
+        visited = {beginWord}
+
+        while queue:
+            word, level = queue.popleft()
+
+            if word == endWord:
+                return level
+
+            for i in range(len(word)):
+                for c in string.ascii_lowercase:
+                    next_word = word[:i] + c + word[i + 1 :]
+
+                    if next_word in wordList and next_word not in visited:
+                        visited.add(next_word)
+                        queue.append((next_word, level + 1))
+
+        return 0
+