<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]:
        res = []
        
        for w in words:
            if self.exist(board, w):
                res.append(w)
        
        return res
    
    def exist(self, board: List[List[str]], word: str) -> bool:
        ROWS, COLS = len(board), len(board[0])
        path = set()

        def dfs(r,c,i):
            if i == len(word):
                return True
            
            if ((r,c) in path or
                min(r,c) < 0 or 
                r >= ROWS or
                c >= COLS or
                board[r][c] != word[i] 
            ):
                return False
            path.add((r,c))
            ans = (
                dfs(r+1,c,i+1) or
                dfs(r,c+1,i+1) or
                dfs(r-1,c,i+1) or
                dfs(r,c-1,i+1)
            )
            path.remove((r,c))
            return ans

            

        # To prevent TLE,reverse the word if frequency of the first letter is more than the last letter's
        count = defaultdict(int, sum(map(Counter, board), Counter()))
        if count[word[0]] > count[word[-1]]:
            word = word[::-1]
            
        for r in range(ROWS):
            for c in range(COLS):
                if dfs(r, c, 0):
                    return True
        return False
16400