<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) {
    map<char, vector<pair<int, int>>> m;

    for (int i = 0; i < board.size(); i++) {
      for (int j = 0; j < board[i].size(); j++) {
        if (m.count(board[i][j]) == 0) {
          m[board[i][j]] = vector<pair<int, int>>{};
        }

        m[board[i][j]].push_back({i, j});
      }
    }

    map<char, int> cnt;

    for (auto e: word) {
      cnt[e]++;
    }

    for (auto e: cnt) {
      if (m[e.first].size() < e.second) {
        return false;
      }
    }
  
    set<pair<int, int>> visited;

    // prevent self referencing error
    function<bool(int, pair<int,int>)> DFS;

    DFS = [&m, &visited, &DFS, word](int depth, pair<int, int> pos) {
      if (depth == word.length()) {
        return true;
      }

      visited.insert(pos);

      bool res = false;
      for (auto e : m[word[depth]]) {
        if ((abs(e.first - pos.first) + abs(e.second - pos.second) <= 1) && (e != pos) && (visited.count(e) == 0)) {
          res |= DFS(depth + 1, e);
        }
      }

      visited.erase(pos);

      return res;
    };

    for (auto e : m[word[0]]) {
      if (DFS(1, e)) {
        return true;
      }
    }

    return false;
  }
};
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