<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.next = {}
        self.word = None

class Solution:
    def buildTrie(self, board, m, n, words: List[str]) -> Trie:
        counter = defaultdict(int)
        for i in range(m):
            for j in range(n):
                counter[board[i][j]] += 1
        rootTrie = Trie()
        for word in words:
            if word[0] not in counter:
                continue
            trie = rootTrie
            for char in word:
                if char not in trie.next:
                    trie.next[char] = Trie()
                trie = trie.next[char]
            trie.word = word
        return rootTrie


    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        m = len(board)
        n = len(board[0])
        trie = self.buildTrie(board, m, n, words)
        foundWords = []
        for i in range(m):
            for j in range(n):
                self.dfs(board, i, j, m, n, trie, foundWords)
        return foundWords

    def dfs(self, board: List[List[str]], i, j, m, n, trie: Trie, foundWords):
        char = board[i][j]
        if char == '\0':
            return

        if char not in trie.next:
            return

        nextTrie = trie.next[char]
        if nextTrie.word is not None:
            foundWords.append(nextTrie.word)
            nextTrie.word = None

        board[i][j] = '\0'
        if i > 0:
            self.dfs(board, i - 1, j, m, n, nextTrie, foundWords)
        if j > 0:
            self.dfs(board, i, j - 1, m, n, nextTrie, foundWords)
        if i + 1 < m:
            self.dfs(board, i + 1, j, m, n, nextTrie, foundWords)
        if j + 1 < n:
            self.dfs(board, i, j + 1, m, n, nextTrie, foundWords)
        board[i][j] = char

        if len(nextTrie.next) == 0:
            del trie.next[char]
            return

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