<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>

// loop through half of the word, if at any point, the current char appears more frequently than the char on the opposite side of the word, reverse it
// catch obvious cases: if the length of the word is larger than the number of chars on the board, or if the frequency of any character in the word is less than 1 (decrementing with each char)
// dont pass a string into the dfs, just use the current index
// instead of using a vis set, update the actual board before and after going deeper

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        if (board[0].empty()) return false;

        const int m = board.size();
        const int n = board[0].size();
        unordered_map<char, int> freq;

        // for (int i = 0; i < m; i++) {
        //     for (int j = 0; j < n; j++) {
        //         cout << board[i][j] << " ";
        //     }
        //     cout << endl;
        // }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                freq[board[i][j]]++;
            }
        }

        if (freq[word.size() - 1] < freq[word[0]]) reverse(word.begin(), word.end());

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] != word[0]) continue;
                // cout << endl << i << " " << j << endl;
                set<pair<int, int>> v;
                if (traverse(board, word, v, make_pair(i, j), m, n, {word[0]})) return true;
            }
        }

        return false;
    }

    bool traverse(const vector<vector<char>>& board, const string& word, set<pair<int, int>>& vis, pair<int, int> cur, const int& m, const int& n, string s) {
        // cout << cur.first << " " << cur.second << " " << s << endl;
        vis.insert(cur);
        if (s == word) return true;
        for (pair<int, int> next: neigh(cur.first, cur.second, m, n)) {
            // cout << "\t" << next.first << " " << next.second << endl;
            if (board[next.first][next.second] != word[s.size()]) continue;
            if (vis.find(next) != vis.end()) continue;
            if (traverse(board, word, vis, next, m, n, s + board[next.first][next.second])) return true;
        }
        vis.erase(cur);
        return false;
    }

    vector<pair<int, int>> neigh(const int& r, const int& c, const int& m, const int& n) {
        vector<pair<int, int>> neighs;
        if (r-1 >= 0) neighs.push_back(make_pair(r-1, c));
        if (r+1 < m) neighs.push_back(make_pair(r+1, c));
        if (c-1 >= 0) neighs.push_back(make_pair(r, c-1));
        if (c+1 < n) neighs.push_back(make_pair(r, c+1));
        return neighs;
    }
};
201956