<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 Trie:
    def __init__(self):
        self.children = {}
        self.end = False
        self.freq = 0

    def insert(self, word):
        curr = self
        curr.freq += 1
        for c in word:
            if c not in curr.children:
                curr.children[c] = Trie()
            curr = curr.children[c]
            curr.freq += 1
        curr.end = True
    
    def remove(self, word):
        curr = self
        curr.freq -= 1
        for c in word:
            curr = curr.children[c]
            curr.freq -= 1
        curr.end = False

class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        trie = Trie()
        for word in words:
            trie.insert(word)
        
        res = []

        def backtrack(r, c, currNode, currWord, hs):
            if currNode.end == True:
                trie.remove(currWord)
                currNode.end = False
                res.append(currWord)
            
            if r not in range(len(board)) or c not in range(len(board[0])) or (r,c) in hs or board[r][c] not in currNode.children:
                return
            if currNode.children[board[r][c]].freq == 0:
                return
            
            currWord += board[r][c]
            currNode = currNode.children[board[r][c]]
            hs.add((r,c))
            backtrack(r+1, c, currNode, currWord, hs)
            backtrack(r-1, c, currNode, currWord, hs)
            backtrack(r, c+1, currNode, currWord, hs)
            backtrack(r, c-1, currNode, currWord, hs)
            hs.discard((r,c))


        for r in range(len(board)):
            for c in range(len(board[0])):
                backtrack(r, c, trie, "", set())

        return res

        
20000