<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>

// fm-vo
class Node {
public:
    Node* child[26] = {NULL};
    bool isEndOfWord;
    Node(){
        this->isEndOfWord = false;
    }
};

class Trie {
public:
    Node* head;
    Trie(){
        head = new Node();
    }

    void addWord(string st) {
        Node* temp = head;
        for(auto i: st) {
            if(!temp->child[i-'a'] ) {
                Node* node = new Node();
                temp->child[i-'a'] = node;
            }
            temp = temp->child[i-'a'];
        }
        
    }

    bool isWordPresent(string st) {
        Node* temp = head;
        for(auto i: st) {
            if(!temp->child[i-'a'] ) {
                return false;
            }
            temp = temp->child[i-'a'];
        }
        
        return true;
    }
};



class Solution {
public: 
    
    void dfs(int x, int y, vector<vector<int>> &vis, vector<vector<char>> &board, string cur, Trie* trie, unordered_map<string, int> &mp, vector<string> &ans) {
        if(!trie->isWordPresent(cur)) {
            return;
        }
        if(mp[cur]) {
            mp[cur]--;
            ans.push_back(cur);
        }
        vis[x][y] = 1;
        int dx[4] = {0, 0, 1, -1};
        int dy[4] = {1, -1, 0, 0};
        for(int i=0; i<4; i++) {
            int new_x = x + dx[i];
            int new_y = y + dy[i];
            if(new_x >=0 && new_y >= 0 && new_x<board.size() && new_y < board[0].size() && vis[new_x][new_y] == 0) {
                dfs(new_x, new_y, vis, board, cur + board[new_x][new_y], trie, mp, ans);
            }
        }
        vis[x][y] = 0;
        
    }
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        Trie* trie = new Trie();
        int n = board.size();
        int m = board[0].size();
        unordered_map<string, int> mp;
        for(auto i: words) {
            trie->addWord(i);
            mp[i] = 1;
        }

        vector<string> ans;
        vector<vector<int>> vis(n, vector<int>(m, 0));
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                string cur = "";
                dfs(i, j, vis, board, cur + board[i][j], trie, mp, ans);
            }
        }

        return ans;
        
    }
};
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