<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:
    bool exist(vector<vector<char>>& board, string word) {
        int ROWS = board.size();
        int COLS = board[0].size();
        bool res = false;
        set<pair<int,int>> path;
        function <bool(int, int, int)> backtrack = [&] (int wordIdx, int row, int col) {
            if (wordIdx == word.size())
                return true;
            if (row < 0 || row >= ROWS ||
                col < 0 || col >= COLS ||
                word[wordIdx] != board[row][col] ||
                path.find(pair<int,int>(row,col)) != path.end())
                return false;
            
            path.insert(pair<int,int>(row,col));
            res = backtrack(wordIdx + 1, row + 1, col) ||
                    backtrack(wordIdx + 1, row - 1, col) ||
                    backtrack(wordIdx + 1, row, col + 1) ||
                    backtrack(wordIdx + 1, row, col - 1);
            path.erase(pair<int,int>(row,col));
            return res;
        };
        
        if (word.size() > ROWS * COLS)
            return false;

        for (char ch : word) {
            bool found = false;
            for (int r = 0; r < ROWS; r++) {
                for (int c = 0; c < COLS; c++) {
                    if (ch == board[r][c]) {
                        found = true;
                        break;
                    }
                }
                if (found)
                    break;
            }
            if (!found)
                return false;
        }

        for (int r = 0; r < ROWS; r++) {
            for (int c = 0; c < COLS; c++) {
                if (board[r][c] == word[0])
                    if (backtrack(0, r, c))
                        return true;
            }
        }
        return false;
    }
};
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