<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:
                node=node.setdefault(letter,{})
            node[word_key]=word
        rows=len(board)
        columns=len(board[0])
        matched=[]
        def backtracking(row,col,parent):
            letter=board[row][col]
            currNode=parent[letter]
            word_match=currNode.pop(word_key,False)
            if word_match:
                matched.append(word_match)
            board[row][col]='#'
            for ro,co in [(-1,0),(0,1),(1,0),(0,-1)]:
                rnew,cnew=row+ro,col+co
                if rnew<0 or rnew>=rows or cnew<0 or cnew>=columns:
                    continue
                if not board[rnew][cnew] in currNode:
                    continue
                backtracking(rnew,cnew,currNode)
            board[row][col]=letter
            if not currNode:
                parent.pop(letter)
        for row in range(rows):
            for col in range(columns):
                if board[row][col] in trie:
                    backtracking(row,col,trie)
        return matched

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