<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 {
    struct Node {
        string word;
        vector<Node*> next;
        Node() : word(""), next(26, nullptr) {}
    };
public:
    Node* root = new Node();
    vector<string> ans;

    void insert(const string& word) {
        Node* p = this->root;
        for(char c : word) {
            if (p->next[c - 'a'] == nullptr) {
                p->next[c - 'a'] = new Node();
            } 
            p = p->next[c - 'a'];
        }
        p->word = word;
    }

    vector<vector<int>> dirs = {{0,1}, {0,-1}, {1,0}, {-1,0}};

    void dfs(vector<vector<char>>& board, int r, int c, Node* p) {
        char letter = board[r][c];
        Node* q = p->next[letter - 'a'];

        if (q->word != "") {
            ans.push_back(q->word);
            q->word = ""; // delete word from trie once found
        }


        board[r][c] = '#';
        for(auto dir : dirs) {
            int nr = r + dir[0];
            int nc = c + dir[1];
            if (nr < 0 || nr >= board.size() || nc < 0 || nc >= board[0].size() || board[nr][nc] == '#') continue;
            if (q->next[board[nr][nc] - 'a'] == nullptr) continue;
            dfs(board, nr, nc, q);
        }
        board[r][c] = letter;

        for(int i = 0; i < 26; i++) {
            if (q->next[i] != nullptr) return;
        }
        p->next[letter-'a'] = nullptr;
    }

    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        // O(len(words) * log(10)) to insert all words into a trie
        for(const string& word : words) insert(word);
        // search: dfs the board
        for(int r = 0; r < board.size(); r++) {
            for(int c = 0; c < board[0].size(); c++) {
                if (this->root->next[board[r][c] - 'a'] != nullptr) 
                    dfs(board, r, c, this->root);
            }
        }
        return ans;
    }
};
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