<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 -> s<sub>1</sub> -> s<sub>2</sub> -> ... -> 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 <= i <= 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> </p> <p><strong>Example 1:</strong></p> <pre> <strong>Input:</strong> beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] <strong>Output:</strong> 5 <strong>Explanation:</strong> One shortest transformation sequence is "hit" -> "hot" -> "dot" -> "dog" -> cog", which is 5 words long. </pre> <p><strong>Example 2:</strong></p> <pre> <strong>Input:</strong> beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"] <strong>Output:</strong> 0 <strong>Explanation:</strong> The endWord "cog" is not in wordList, therefore there is no valid transformation sequence. </pre> <p> </p> <p><strong>Constraints:</strong></p> <ul> <li><code>1 <= beginWord.length <= 10</code></li> <li><code>endWord.length == beginWord.length</code></li> <li><code>1 <= wordList.length <= 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>