<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>

from collections import defaultdict
from functools import reduce
from operator import getitem


class Trie(defaultdict):
    def __init__(self, words=None):
        super().__init__(Trie)
        self.is_word = False
        self.count = 0

        if words:
            for word in words:
                self.add(word)

    
    def add(self, word):
        for c in word:
            (self:=self[c])
        self.is_word = True

    def search(self, word):
        return reduce(getitem, word, self).is_word
    
class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        M, N = len(board), len(board[0])
        L = len(words)
        K = len(max(words, key=len))
        res = []
        visited = set()
        def backtrack(x, y, trie, word):
            if trie.is_word:
                res.append(word)
                trie.is_word=False
            if len(word) == K:
                return
            if len(res) == L:
                return

            for nx, ny in (x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1):
                if 0 <= nx < M and 0 <= ny < N and (nx, ny) not in visited:
                    char = board[nx][ny]
                    if char in trie:
                        visited.add((nx, ny))
                        backtrack(nx, ny, trie[char], word + char)
                        visited.remove((nx, ny))


        trie = Trie(words)

        for i in range(M):
            for j in range(N):
                visited.add((i, j))
                backtrack(i, j, trie[board[i][j]], board[i][j])
                visited.remove((i, j))

        return res

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