<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]:
        # trie
        # build trie
        trie = defaultdict(list)
        for word in words:
            node = trie
            for char in word:
                if char not in node:
                    node[char] = defaultdict(list)
                node = node[char]
            # mark end
            node["END"] = True

        # search word
        def search(row, col, node, cur_word):
            if "END" in node:
                res.add(cur_word)
                del node["END"]  # Prune the trie to avoid duplicate work
            if row < 0 or row >= m or col < 0 or col >= n or visited[row][col]:
                return
            char = board[row][col]
            if char not in node:
                return 
            # search child
            visited[row][col] = True
            directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
            for r , c in directions:
                nr, nc = row + r, col + c
                search(nr, nc, node[char], cur_word+char)
            # backtrack
            visited[row][col] = False

        m, n = len(board), len(board[0])
        res = set()
        visited = [[False] * n for _ in range(m)]
        for i in range(m):
            for j in range(n):
                if board[i][j] in trie:
                    search(i, j, trie, "")

        return res
17400