<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) {
        if (board[0][0] == word[0]) 
            reverse(word.begin(), word.end());
        int m = board.size();
        int n = board[0].size();
        
        if (word.length() > n * m) return false; // Optimization check

        set<pair<int, int>> visited;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (verification(i, j, 0, board, word, visited)) {
                    return true;
                }
            }
        }
        return false;
    }

private:
    bool verification(int i, int j, int index, vector<vector<char>>& board, string& word, set<pair<int, int>>& visited) {
        // Check bounds and visited status
        if (i < 0 || i >= board.size() || j < 0 || j >= board[0].size() || visited.count({i, j}) || board[i][j] != word[index]) {
            return false;
        }
        
        // If we have matched all characters of the word
        if (index == word.length() - 1) {
            return true;
        }
        
        // Mark the cell as visited
        visited.insert({i, j});
        
        // Explore all neighbors
        vector<vector<int>> neigh = neighbors(i, j, board);
        for (const auto& cell : neigh) {
            int ni = cell[0];
            int nj = cell[1];
            if (verification(ni, nj, index + 1, board, word, visited)) {
                return true;
            }
        }
        
        // Backtrack: Unmark the cell as visited
        visited.erase({i, j});
        return false;
    }

    vector<vector<int>> neighbors(int i, int j, vector<vector<char>>& board) {
        int m = board.size();
        int n = board[0].size();
        vector<vector<int>> result;
        
        if (i > 0) result.push_back({i - 1, j}); // Up
        if (i < m - 1) result.push_back({i + 1, j}); // Down
        if (j > 0) result.push_back({i, j - 1}); // Left
        if (j < n - 1) result.push_back({i, j + 1}); // Right
        
        return result;
    }
};
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