<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]:
        WORD_KEY = "$"

        trie = {}
        for word in words:
            node = trie
            for letter in word:
                if letter not in node:
                    node[letter] = {}
                
                node = node[letter]

            node[WORD_KEY] = word
        
        matches = []
        m = len(board)
        n = len(board[0])

        def backtrack(row, col, trie, visited):
            l = board[row][col]

            if l in trie:
                currNode = trie[l]

                if WORD_KEY in currNode:
                    matches.append(currNode[WORD_KEY])
                    del currNode[WORD_KEY]
                
                adj = [[0,1],[0,-1],[1,0],[-1,0]]
                visited.add((row,col))

                for mv in adj:
                    nrow = row + mv[0]
                    ncol = col + mv[1]

                    if (nrow,ncol) not in visited and 0 <= nrow < m and 0 <= ncol < n:
                        backtrack(nrow, ncol, currNode, visited)
                
                visited.remove((row,col))
            
                if not trie[l]:
                    del trie[l]
        
        for row in range(m):
            for col in range(n):
                backtrack(row, col, trie, set())
        
        return matches
17500