<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
class Solution:
    
    def find(self, word, i, j):
        if len(word) == 0:
            return True
        if (not 0 <= i < len(self.board)) or (not 0 <= j < len(self.board[0])) or word[0] != self.board[i][j] or self.visited[i][j]:
            return False
        self.visited[i][j] = True
        ans = self.find(word[1:], i + 1, j) or self.find(word[1:], i, j + 1) or self.find(word[1:], i - 1, j) or self.find(word[1:], i, j - 1)
        self.visited[i][j] = False
        return ans
    
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        self.board = board
        self.visited = [[False for _ in range(len(board[0]))] for _ in range(len(board))]
        self.m = defaultdict(list)
        for i in range(len(board)):
            for j in range(len(board[0])):
                self.m[self.board[i][j]].append((i, j))
        ans = []
        reverse = len(set(word[:len(word) // 2] for word in words)) < 2
        for word in words:
            word = word[::-1] if reverse else word
            for i, j in self.m[word[0]]:
                if self.find(word, i, j):
                    ans.append(word[::-1] if reverse else word)
                    break
        return ans
16900