<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 Trie:
    def __init__(self, word = None, is_end = False):
        self.word = word
        self.is_end = is_end
        self.children = {}

    def add(self, word :str):
        current = self
        for letter in word:
            if letter not in current.children:
                current.children[letter] = Trie()
            current = current.children[letter]
        current.is_end = True
        current.word = word


class Solution:
    directions = [(1,0), (0,1), (-1, 0), (0,-1)]
    rows = None
    columns = None
    trie = None
    result = set()

    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        self.board = board
        self.rows = len(self.board)
        if self.rows < 1:
            return result
        self.columns = len(self.board[0])
        self.trie = Trie()
        self.result = set()

        for word in words:
            self.trie.add(word)
        
        for row in range(self.rows):
            for column in range(self.columns):
                if self.board[row][column] in self.trie.children:
                    self.backtrack(row, column, self.trie)

        return list(self.result)

    def backtrack(self, row:int, column: int, current_node: Trie):
        if current_node.is_end:
            self.result.add(current_node.word)

        if not self.isValid(row, column):
            return

        letter = self.board[row][column]
        if letter not in current_node.children:
            return
        
        trie = current_node.children[letter]
        self.board[row][column] = ""
        for (x,y) in self.directions:
            new_row = row + x
            new_column = column + y
            self.backtrack(new_row, new_column, trie)
        
        self.board[row][column] = letter
    
    def isValid(self, row: int, column: int) -> bool:
        if row < 0 or column < 0 or row >= self.rows or column >= self.columns:
            return False
        return True

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