<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 check(vector<vector<char>>& board, string& word, int index, int row, int col) {
        if (index == word.size()) {  return true; }

        vector<pair<int, int>> directions = {{0,1}, {0,-1}, {1,0}, {-1,0}};
        for (pair<int, int> direction : directions) {
            int newRow = row + direction.first;
            int newCol = col + direction.second;
            if (newRow >= 0 && newRow < board.size() && newCol >= 0 && newCol < board[0].size() && board[newRow][newCol] == word[index]) {
                board[newRow][newCol] = '0';
                if (check(board, word, index + 1, newRow, newCol)) {
                    return true;
                }
                board[newRow][newCol] = word[index];
            }
        }
        return false;
    }

        bool exist(vector<vector<char>>& board, string word) {

            // Prune unnecessary checks
        vector<int> char_count(256, 0);
        for (const auto& row : board) {
            for (const char c : row) {
                char_count[c]++;
            }
        }
        for (const char c : word) {
            if (char_count[c]-- <= 0) return false;
        }

        for (int i = 0; i < board.size(); i++) {
            for (int j = 0; j < board[0].size(); j++) {
                if (board[i][j] == word[0]) {
                    board[i][j] = '0';
                    if (check(board, word, 1, i, j)) {
                        return true;
                    }
                    board[i][j] = word[0];
                }
            }
        }
        return false;
    }
};

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