<p>Given an <code>m x n</code> grid of characters <code>board</code> and a string <code>word</code>, return <code>true</code> <em>if</em> <code>word</code> <em>exists in the grid</em>.</p>

<p>The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/04/word2.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> board = [[&quot;A&quot;,&quot;B&quot;,&quot;C&quot;,&quot;E&quot;],[&quot;S&quot;,&quot;F&quot;,&quot;C&quot;,&quot;S&quot;],[&quot;A&quot;,&quot;D&quot;,&quot;E&quot;,&quot;E&quot;]], word = &quot;ABCCED&quot;
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/04/word-1.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> board = [[&quot;A&quot;,&quot;B&quot;,&quot;C&quot;,&quot;E&quot;],[&quot;S&quot;,&quot;F&quot;,&quot;C&quot;,&quot;S&quot;],[&quot;A&quot;,&quot;D&quot;,&quot;E&quot;,&quot;E&quot;]], word = &quot;SEE&quot;
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/15/word3.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> board = [[&quot;A&quot;,&quot;B&quot;,&quot;C&quot;,&quot;E&quot;],[&quot;S&quot;,&quot;F&quot;,&quot;C&quot;,&quot;S&quot;],[&quot;A&quot;,&quot;D&quot;,&quot;E&quot;,&quot;E&quot;]], word = &quot;ABCB&quot;
<strong>Output:</strong> false
</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;= 6</code></li>
	<li><code>1 &lt;= word.length &lt;= 15</code></li>
	<li><code>board</code> and <code>word</code> consists of only lowercase and uppercase English letters.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you use search pruning to make your solution faster with a larger <code>board</code>?</p>

class Solution {
public:

    struct Node
    {
        Node(){}
        unordered_map<char, Node*> child;  
    };
    
    Solution():dr{0, 0, -1, 1}, dc{1, -1, 0, 0}
    {
        root = new Node();
    }
    
    void build_trie(vector<vector<char>> &board, int r, int c, int len, Node *node)
    {
        if(len == 0)
            return;        
        char ch = board[r][c] ;
        board[r][c] = '#';        

        if(!node->child[ch])
        {
            node->child[ch] = new Node();
        }
        node = node->child[ch];

        for( int i = 0; i < dr.size(); i++)
        {
            int rr = dr[i] + r;
            int cc = dc[i] + c;
            if(rr < 0 || cc < 0 || rr >= board.size() || cc >= board[0].size() || board[rr][cc] == '#')
                continue;
            build_trie(board, rr, cc, len-1, node);
        }
        
        board[r][c] = ch;
    }
    
    bool search_trie(string &word)
    {
        Node *node = root;
        for(int i = 0; i < word.size(); i++)
        {
            if(!node->child[word[i]])
                return false;
            node = node->child[word[i]];
        }
        return true;
    }
    
    bool backtrack(vector<vector<char>> &board, int r, int c, string &word, int index)
    {
        if(index >= word.size())
            return true;
        char ch = board[r][c];
        board[r][c] = '#';
        
        for( int i = 0; i < dr.size(); i++)
        {
            int rr = r + dr[i];
            int cc = c + dc[i];
            if(rr < 0 || cc < 0 || rr >= board.size() || cc >= board[0].size() || 
              board[rr][cc] != word[index])
                continue;
            if( backtrack(board, rr, cc, word, index+1))
                return true;
        }
        
        board[r][c] = ch;
        return false;
    }
    
    bool exist(vector<vector<char>>& board, string word) {
        int index = 0;
        for(int r = 0; r < board.size(); r++)
            for(int c = 0; c < board[0].size(); c++)
            {
                if(board[r][c] == word[index])
                {
                    // if(backtrack(board, r, c, word, index+1))
                        // return true;
                    build_trie(board, r, c, word.size(), root);
                }
            }
        return search_trie(word);
        // return false;
    }
    
private:
    vector<int> dr;
    vector<int> dc;
    Node *root;
};
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