<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 isVisited(vector<pair<int, int>> &visited, pair<int, int> p) {
    for (const auto &pt : visited)
      if (pt.first == p.first && pt.second == p.second)
        return true;
    return false;
  }
  vector<pair<int, int>> searchAdj(pair<int, int> s, char t,
                                   vector<vector<char>> &board,
                                   vector<pair<int, int>> &visited) {
    vector<pair<int, int>> res;
    if (s.first < 0 || s.second < 0) {
      for (int i = 0; i < board.size(); i++) {
        for (int j = 0; j < board[0].size(); j++) {
          if (board[i][j] == t)
            res.push_back({i, j});
        }
      }
      return res;
    } else {
      vector<int> p = {-1, 1};
      for (auto i : p) {
        if (s.first + i < board.size() && board[s.first + i][s.second] == t &&
            !isVisited(visited, {s.first + i, s.second}))
          res.push_back({s.first + i, s.second});
        if (s.second + i < board[0].size() &&
            board[s.first][s.second + i] == t &&
            !isVisited(visited, {s.first, s.second + i}))
          res.push_back({s.first, s.second + i});
      }
      return res;
    }
  }
  bool search(vector<vector<char>> &board, int ind, pair<int, int> &curr,
              string word, vector<pair<int, int>> &visited) {
    if (ind >= word.length())
      return true;
    vector<pair<int, int>> points = searchAdj(curr, word[ind], board, visited);
    bool f = false;
    for (auto &pt : points) {
      visited.push_back(pt);
      f = search(board, ind + 1, pt, word, visited);
      visited.pop_back();
      if (f)
        return true;
    }
    return false;
  }
  bool exist(vector<vector<char>> &board, string word) {
    pair<int, int> curr = {-1, -1};
    vector<pair<int, int>> visited;
    for (int i = 0; i < word.size() / 2; i++) {
      char t = word[i];
      word[i] = word[word.size() - i - 1];
      word[word.size() - i - 1] = t;
    }
    return search(board, 0, curr, word, visited);
  }
};
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