<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 TrieNode:
    def __init__(self):
        self.isEnd = False
        self.arr = [None for i in range(26)]


class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        output = set()
        start = TrieNode()

        def helper(x, y, trie, result):
            if x < 0 or y < 0 or x >= len(board) or y >= len(board[0]) or board[x][y] == -1:
                return
            if not trie.arr[ord(board[x][y]) - ord("a")]:
                return
            if trie.arr[ord(board[x][y]) - ord("a")] and trie.arr[ord(board[x][y]) - ord("a")].isEnd:
                output.add(result+board[x][y])
            temp = board[x][y]
            board[x][y] = -1
            helper(x, y+1, trie.arr[ord(temp) - ord("a")], result+temp)
            helper(x, y-1, trie.arr[ord(temp) - ord("a")], result+temp)
            helper(x+1, y, trie.arr[ord(temp) - ord("a")], result+temp)
            helper(x-1, y, trie.arr[ord(temp) - ord("a")], result+temp)
            board[x][y] = temp
            return
        
        for i in words:
            current = start
            for j in range(len(i)):
                if current.arr[ord(i[j])-ord("a")]:
                    current = current.arr[ord(i[j])-ord("a")]
                else:
                    newTrieNode = TrieNode()
                    current.arr[ord(i[j])-ord("a")] = newTrieNode
                    current = current.arr[ord(i[j])-ord("a")]
                if j == len(i) - 1:
                    current.isEnd = True
        for i in range(len(board)):
            for j in range(len(board[0])):
                helper(i, j, start, "")
        return list(output)

            
            

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