<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 {
private:
    static const vector<vector<int>> dirs_;
public:
    bool dfs(vector<vector<char>>& board, string& word, vector<vector<bool>>& seen,  vector<int> start, int i) {
        int x = start[0];
        int y = start[1];
        if(board[x][y] != word[i]) return false;
        if(board[x][y] == word[i] && i == word.size() - 1) return true;
        seen[x][y] = true;

        // cout << board[x][y] << ", " << word[i] << endl;

        for(auto& dir:dirs_) {
            int new_x = x + dir[0];
            int new_y = y + dir[1];
            if(new_x >= 0 && new_x < board.size()) {
                if(new_y >= 0 && new_y < board[0].size()) {
                    if(!seen[new_x][new_y]) {
                        if(dfs(board, word, seen, {new_x, new_y}, i+1)) {
                            return true;
                        }
                    }
                }
            }
        } 
        seen[x][y] = false;
        return false;
    }

    bool exist(vector<vector<char>>& board, string word) {
        int m = board.size();
        int n = board[0].size();

        unordered_map<char, bool> exists;
        for(int i=0; i<m; ++i) {
            for(int j=0; j<n; ++j) {
                exists[board[i][j]] = true;
            }
        }
        for(auto& c: word) {
            if(!exists[c]) return false;
        }

        vector<vector<bool>> seen(m, vector<bool>(n));
        for(int i=0; i<m; ++i) {
            for(int j=0; j<n; ++j) {
                if(dfs(board, word, seen, {i, j}, 0)) return true;
            }
        }
        return false;
    }
};

const vector<vector<int>> Solution::dirs_ = {{0,1}, {0,-1}, {1,0}, {-1,0}};
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