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

        unordered_map<char, int> freq;
        for(int row=0; row < m; row++){
            for(int col=0; col < n; col++){
                char curr = board[row][col];
                auto found = freq.find(curr);
                if(found == freq.end()){
                    freq.insert({curr, 1});
                } else{
                    found->second += 1;
                }
            }
        }

        auto last = freq.find(word[word.length() - 1]);
        auto first = freq.find(word[0]);
        if(last == freq.end() || first == freq.end()){
            return false;
        }
        if(last->second > first->second){
            reverse(word.begin(), word.end());
        }

        for(int row=0; row < m; row++){
            for(int col=0; col < n; col++){
                unordered_set<int> memory;
                bool result = backtrack(board, memory, word, 0, {row, col});
                if(result){ return true; }
            }
        }
        
        return false;
    }

private:
    bool backtrack(
        vector<vector<char>>& board,
        unordered_set<int>& memory,
        string& word,
        int index,
        pair<int, int> curr
    ){
        if(index == word.length()){
            return true;
        }

        char correct = word[index];

        int row = curr.first;
        int col = curr.second;
        int rep = row*10 + col;

        if(row < 0 || row >= m || col < 0 || col >= n){
            return false;
        }

        if(board[row][col] == correct && memory.find(rep) == memory.end()){
            memory.insert(rep);
            bool up = backtrack(board, memory, word, index+1, {row - 1, col});
            bool down = backtrack(board, memory, word, index+1, {row + 1, col});
            bool left = backtrack(board, memory, word, index+1, {row, col - 1});
            bool right = backtrack(board, memory, word, index+1, {row, col + 1});
            memory.erase(rep);
            if(up || down || left || right){
                return true;
            }
        }
        return false;
    }
};
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