<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:

    const vector<pair<int,int>> dirs = {{-1,0}, {1,0}, {0,-1}, {0,1}};

    bool ok(vector<vector<char>>& board, pair<int,int>& pos)
    {
        int m = board.size();
        int n = board[0].size();
        return (pos.first >= 0 && pos.first < m && pos.second >= 0 && pos.second < n);
    }

    bool rec(vector<vector<char>>& board, string& word, string cur, pair<int,int>& pos, vector<vector<bool>>& visited)
    {
        if (cur == word)
            return true;
        // cout << cur << endl;

        visited[pos.first][pos.second] = true;
        for(const auto& dir: dirs)
        {
            pair<int,int> newPos = {pos.first + dir.first, pos.second + dir.second};
            if (ok(board, newPos) && !visited[newPos.first][newPos.second])
            {
                string shrikara = cur + board[newPos.first][newPos.second];
                if ((word.size() >= shrikara.size()) && (shrikara[shrikara.size()-1] == word[shrikara.size()-1]))
                {
                    if (rec(board, word, cur + board[newPos.first][newPos.second], newPos, visited))
                        return true;
                    visited[newPos.first][newPos.second] = false;
                }
            }
        }

        return false;
    }

    bool exist(vector<vector<char>>& board, string word) {
        int m = board.size();
        int n = board[0].size();

        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < n; j++)
            {
                // cout << "Start: " << i << " " << j << endl;
                pair<int,int> p = make_pair(i,j);
                vector<vector<bool>> starter(m, vector<bool>(n, false));
                if (rec(board, word, {board[i][j]}, p, starter))
                    return true;
            }
        }

        return false;
    }
};
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