<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 Trie:
    def __init__(self):
        self.root = [None]*27

    def addWord(self, word, word_idx):
        curr = self.root
        for ch in word:
            key = ord(ch)-96
            if not curr[key]:
                curr[key] = [None]*27
            curr = curr[key]
        curr[0] = word_idx

class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        trie = Trie()
        for idx, word in enumerate(words):
            trie.addWord(word, idx)
        opt = set()
        neighbors = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        M, N = len(board), len(board[0])
        def dfs(board, node, visited, row, col, M, N):
            if node[0] is not None:
                opt.add(node[0])
            for r, c in neighbors:
                row_new, col_new = row+r, col+c
                if 0 <= row_new < M and 0 <= col_new < N and (row_new, col_new) not in visited:
                    ch = board[row_new][col_new]
                    key = ord(ch)-96
                    if node[key]:
                        visited.add((row_new, col_new))
                        dfs(board, node[key], visited, row_new, col_new, M, N)
                        visited.remove((row_new, col_new))
        for rdx in range(M):
            for cdx in range(N):
                node = trie.root
                key = ord(board[rdx][cdx])-96
                if node[key] is not None:
                    dfs(board, node[key], {(rdx, cdx)}, rdx, cdx, M, N)
        return [words[widx] for widx in opt]

19000