<p>Given an <code>m x n</code> <code>board</code>&nbsp;of characters and a list of strings <code>words</code>, return <em>all words on the board</em>.</p>

<p>Each word must be constructed from letters of sequentially adjacent cells, where <strong>adjacent cells</strong> are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/07/search1.jpg" style="width: 322px; height: 322px;" />
<pre>
<strong>Input:</strong> board = [[&quot;o&quot;,&quot;a&quot;,&quot;a&quot;,&quot;n&quot;],[&quot;e&quot;,&quot;t&quot;,&quot;a&quot;,&quot;e&quot;],[&quot;i&quot;,&quot;h&quot;,&quot;k&quot;,&quot;r&quot;],[&quot;i&quot;,&quot;f&quot;,&quot;l&quot;,&quot;v&quot;]], words = [&quot;oath&quot;,&quot;pea&quot;,&quot;eat&quot;,&quot;rain&quot;]
<strong>Output:</strong> [&quot;eat&quot;,&quot;oath&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/07/search2.jpg" style="width: 162px; height: 162px;" />
<pre>
<strong>Input:</strong> board = [[&quot;a&quot;,&quot;b&quot;],[&quot;c&quot;,&quot;d&quot;]], words = [&quot;abcb&quot;]
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == board.length</code></li>
	<li><code>n == board[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 12</code></li>
	<li><code>board[i][j]</code> is a lowercase English letter.</li>
	<li><code>1 &lt;= words.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
	<li>All the strings of <code>words</code> are unique.</li>
</ul>

class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        # create trie
        Trie = lambda: defaultdict(Trie)
        trie = Trie()
        END = True
        
        for word in words:
            reduce(dict.__getitem__,word,trie)[END] = word
        
        res = set()
        def findstr(i,j,t):
            if END in t:
                res.add(t[END])
                # return
            letter = board[i][j]
            board[i][j] = ""
            if i > 0 and board[i-1][j] in t:
                findstr(i-1,j,t[board[i-1][j]])
            if j>0 and board[i][j-1] in t:
                findstr(i,j-1,t[board[i][j-1]])
            if i < len(board)-1 and board[i+1][j] in t:
                findstr(i+1,j,t[board[i+1][j]])
            if j < len(board[0])-1 and board[i][j+1] in t:
                findstr(i,j+1,t[board[i][j+1]])
            board[i][j] = letter
            
            return 
        
        for i, row in enumerate(board):
            for j, char in enumerate(row):
                if board[i][j] in trie:
                    findstr(i,j,trie[board[i][j]])
        return res
        
17900