<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:


vector<vector<char>> board;
vector<vector<int>> moves = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}};
string word;


bool backtrack(int row, int col, unordered_map<string, bool> &visited, string current, int matchIdx) {

    if(word == current) {
        return true;
    }

    if(matchIdx < 0 or matchIdx > word.size()) {
        return false;
    }

    for(auto move : moves) {
        int newRow = row + move[0];
        int newCol = col + move[1];

        if(newRow>=0 and newCol>=0 and newRow < board.size() and newCol < board[0].size() and word[matchIdx] == board[newRow][newCol]) {

            string rowCol = to_string(newRow) + "," + to_string(newCol);

            if(!visited[rowCol]) {
                visited[rowCol] = true;
                current.push_back(board[newRow][newCol]);
                bool res = backtrack(newRow, newCol, visited, current, matchIdx+1);

                if(res) {
                    return res;
                }

                current.pop_back();
                visited[rowCol] = false;

            }
        }

    }

    return false;
}

    bool exist(vector<vector<char>>& _board, string _word) {
        unordered_map<string, bool> visited;
        board = _board;
        word = _word; 

        unordered_set<char> charset;

        for(auto w: word) {
            charset.insert(w);
        }

        unordered_set<char> boardSet;

        for(int i = 0; i < board.size(); i++) {
            for(int j = 0; j < board[0].size(); j++) {
                boardSet.insert(board[i][j]);
            }
        }


        for(auto ch: charset) {
            if(!boardSet.contains(ch)) {
                return false;
            }
        }

        for(int i = 0; i < board.size(); i++) {
            for(int j = 0; j < board[i].size(); j++) {
                unordered_map<string, bool> visited;

                if(word[0] == board[i][j]) {
                    string rowCol = to_string(i) + "," + to_string(j);
                    visited[rowCol] = true;
                    string current;
                    current.push_back(board[i][j]);
                    bool res =backtrack(i, j, visited, current, 1);
                    if(res) {
                        return res;
                    }
                }

            }
        }

        return false;
    }

};
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