<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(
      const vector<vector<char>> &board,
      const string &word,
      set<pair<int, int>> &visited,
      int string_pos,
      int x, int y
    ) {
      if (x < 0 || x >= board.size()) return false;
      if (y < 0 || y >= board[x].size()) return false;
      if (string_pos >= word.size()) return false;
      if (visited.count({x, y})) return false;
      if (board[x][y] != word[string_pos]) return false;
      visited.emplace(x, y);

      // We matched on this square, and this was the last char. We are done
      if (string_pos == word.size() - 1) return true;

      // Explore adjacent squares
      if (exist(board, word, visited, string_pos + 1, x + 1, y)) return true;
      if (exist(board, word, visited, string_pos + 1, x - 1, y)) return true;
      if (exist(board, word, visited, string_pos + 1, x, y + 1)) return true;
      if (exist(board, word, visited, string_pos + 1, x, y - 1)) return true;

      visited.erase({x, y});
      return false;
    }

    bool exist(vector<vector<char>>& board, string word) {
      // Cheaty test to make sure the board contains all chars in word
      unordered_set<char> board_chars;
      for (int x = 0, m = board.size(); x < m; ++x) {
        for (int y = 0, n = board[0].size(); y < n; ++y) {
          board_chars.insert(board[x][y]);
        }
      }
      for (char c : word)
        if (!board_chars.count(c))
          return false;

      set<pair<int, int>> visited;
      for (int x = 0, m = board.size(); x < m; ++x) {
        for (int y = 0, n = board[0].size(); y < n; ++y) {
          if (exist(board, word, visited, 0, x, y)) {
            return true;
          }
        }
      }

      return false;
    }
};
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