<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 Node:
    def __init__(self):
        self.children = [None]*26
        self.isComplete = False
class Trie:
    def __init__(self):
        self.root = Node()
    def insert(self,w):
        root = self.root
        for ch in w:
            idx = ord(ch)-ord("a")
            if root.children[idx] == None:
                root.children[idx] = Node()
            root = root.children[idx]
        root.isComplete = True
class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        trie = Trie()
        for w in words:
            trie.insert(w)
        n = len(board)
        m = len(board[0])
        dir_x = [0,0,-1,1]
        dir_y = [-1,1,0,0]
        def isValid(i,j):
            return i>=0 and i<n and j>=0 and j<m
        def solve(i,j,node,ans,res):
            if node.isComplete == True:
                ans.append("".join(res))  
                node.isComplete = False
            ch = board[i][j]
            board[i][j] = "*"
            for d in range(4):
                new_x,new_y = dir_x[d]+i,dir_y[d]+j
                if isValid(new_x,new_y) and board[new_x][new_y]!="*":
                    newId = ord(board[new_x][new_y])-ord('a')
                    if node.children[newId] !=None:
                        res.append(board[new_x][new_y])
                        solve(new_x,new_y,node.children[newId],ans,res)
                        res.pop()
            board[i][j] = ch
            

        
        ans = []
        res = []
        for i in range(n):
            for j in range(m):
                root = trie.root
                ch = board[i][j]
                idx = ord(ch)-ord('a')
                if root.children[idx]!=None:
                    root = root.children[idx]
                    res.append(ch)
                    solve(i,j,root,ans,res)
                    res.pop()
        return ans
        
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