<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) {
        
        int c1 = count(word.begin(), word.end(), word[0]);
        int c2 = count(word.begin(), word.end(), word.back());

        if (c1 > c2) {
            reverse(word.begin(), word.end());
        }

        for (int r = 0; r < board.size(); r++) {
            for (int c = 0; c < board[0].size(); c++) {
                if (board[r][c] == word[0]) {
                    set<tuple<int, int>> visited = {};
                    if (dfs(board, word, 0, r, c, visited)) {
                        return true;
                    }
                }
            }
        }
        return false;
    }
    
private:
    vector<vector<int>> dirs = {{0,-1},{-1,0},{0,1},{1,0}};
    bool dfs(vector<vector<char>>& board, string word, int i, int r, int c, set<tuple<int, int>>& visited) {
        tuple<int, int> t = make_tuple(r,c);
        if (visited.find(t) != visited.end()) {
            return false;
        }
        if (i >= word.size()) {
            return true;
        }
        if (r < 0 || r >= board.size() || c < 0 || c >= board[0].size()) {
            return false;
        }
        if (word[i] != board[r][c]) {
            return false;
        }

        visited.insert(t);
        bool ex = false;

        for (vector<int> dir : dirs) {
            ex = ex || dfs(board, word, i+1, r+dir[0], c+dir[1], visited);
        }

        visited.erase(t);

        return ex;

    }
};
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