<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 rows = board.size();
        int cols = board[0].size();//board[0] refers to the first row of the grid, which is a vector<char>.
                    // board[0].size() returns the number of columns (or the number of elements) in that first row.
        
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (dfs(board, word, i, j, 0)) {// dfs is called to perform a depth-first search (DFS) starting from that cell, checking if the word can be formed starting from this position.

                    return true;
                }
            }
        }
        return false;
    }
    //dfs function used above 
private:
    bool dfs(vector<vector<char>>& board, string& word, int i, int j, int index) {//index-the current position in the word that we're matching.
        if (index == word.size()) {
            return true; // All characters in the word are matched
        }
        if (i < 0 || i >= board.size() || j < 0 || j >= board[0].size() || board[i][j] != word[index]) {
            return false; // Out of bounds or character mismatch
        }
        
        char temp = board[i][j];
        board[i][j] = '#'; // Mark the cell as visited
        
        // Check all four possible directions (up, down, left, right)
                bool found = dfs(board, word, i + 1, j, index + 1) || // Move down
                     dfs(board, word, i - 1, j, index + 1) || // Move up
                     dfs(board, word, i, j + 1, index + 1) || // Move right
                     dfs(board, word, i, j - 1, index + 1);   // Move left

        
        board[i][j] = temp; // Backtrack, restore the cell
        return found;
    }
};
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