<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:
    void backtracking(vector<vector<char>>& board, string& word, bool& exist,
                      unordered_set<char*>& checkSet, int checkIdx, int i,
                      int j) {
        if (exist || checkSet.find(&board[i][j]) != checkSet.end())
            return;
        exist = word.size() - 1 == checkIdx;
        checkIdx++;
        checkSet.insert(&board[i][j]);
        if (!exist && j + 1 < board[i].size() &&
            board[i][j + 1] == word[checkIdx]) {
            backtracking(board, word, exist, checkSet, checkIdx, i, j + 1);
        }
        if (!exist && i + 1 < board.size() &&
            board[i + 1][j] == word[checkIdx]) {
            backtracking(board, word, exist, checkSet, checkIdx, i + 1, j);
        }
        if (!exist && j - 1 >= 0 && board[i][j - 1] == word[checkIdx]) {
            backtracking(board, word, exist, checkSet, checkIdx, i, j - 1);
        }
        if (!exist && i - 1 >= 0 && board[i - 1][j] == word[checkIdx]) {
            backtracking(board, word, exist, checkSet, checkIdx, i - 1, j);
        }
        checkSet.erase(&board[i][j]);
    }

    bool exist(vector<vector<char>>& board, string word) {
        bool exist = false;
        std::reverse(word.begin(), word.end());
        for (int i = 0; i < board.size(); i++) {
            for (int j = 0; j < board[i].size(); j++) {
                unordered_set<char*> checkSet;
                if (word[0] == board[i][j]) {
                    backtracking(board, word, exist, checkSet, 0, i, j);
                }
                if (exist) {
                    return exist;
                }
                checkSet.clear();
            }
        }
        return exist;
    }
};
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