<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>

struct TrieNode {
    unordered_map<char, TrieNode*> children;
    bool eofWord=false;
    string word="";
};
class Solution {
    void dfs(vector<vector<char>>& board, vector<string>& res, TrieNode* root, int x, int y){
        if(x<0 || y<0 || x>=board.size() || y>=board[0].size()){
            return;
        }
        
        char c = board[x][y];
        if(root->children.find(c) == root->children.end()){
            return;
        }
        
        TrieNode* curr = root->children[c];
        if(curr->word != ""){
            res.push_back(curr->word);
            curr->word="";
        }
        board[x][y]='0';
        vector<int> xOffset = {1, -1, 0, 0};
        vector<int> yOffset = {0, 0, 1, -1};
        for(int i=0;i<4;i++){
            dfs(board, res, curr, x+xOffset[i], y+yOffset[i]);
        }
        board[x][y] = c;

        // Optimization
        if(curr->children.size() == 0){
            root->children.erase(c);
        }
    }
public:
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        // Construct TrieNode
        int n = words.size();
        TrieNode* root = new TrieNode;

        for(string s: words){
            TrieNode* curr = root;
            for(char c : s){
                if(curr->children.find(c) == curr->children.end()){
                    curr->children[c] = new TrieNode;
                }
                curr = curr->children[c];
            }
            curr->word = s;
            curr->eofWord = true;
        }

        int nRows = board.size();
        int nCols = board[0].size();
        vector<string> res;
        for(int i=0;i<nRows;i++){
            for(int j=0;j<nCols;j++){
                dfs(board, res, root, i, j);
            }
        }
        return res;
    }
};
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