<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.nexts = {}
        self.end = False
    
    def addWord(self, word: str):
        node = self
        for ch in word:
            if ch not in node.nexts:
                node.nexts[ch] = Trie()
            node = node.nexts[ch]
        node.end = True

class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        # need to design data structure to store words in DONE
        # need to search for words starting at each square
        # need to use data structure to terminate search early when can
        ROWS, COLS = len(board), len(board[0])
        DIRS = [[1, 0],[-1, 0],[0, 1],[0, -1]]
        trie = Trie()
        for word in words:
            trie.addWord(word)
        
        res = set()
        letters = []
        path = set()

        def inBounds(r, c):
            return 0 <= r < ROWS and 0 <= c < COLS
        def dfs(r, c, node):
            if node.end:
                res.add("".join(letters))
                node.end = False
            if not inBounds(r, c) or (r, c) in path:
                return
            ch = board[r][c]
            if ch not in node.nexts:
                return

            path.add((r,c))
            letters.append(ch)
            for dr, dc in DIRS:
                row, col = r + dr, c + dc
                dfs(row, col, node.nexts[ch])
            letters.pop()
            path.remove((r,c))


        for r in range(ROWS):
            for c in range(COLS):
                dfs(r, c, trie)
        
        return list(res)


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