LeetCode 2267. Check if There Is a Valid Parentheses String Path

class Solution {
public:
    bool hasValidPath(vector<vector<char>> &grid) {
        int m = grid.size(), n = grid[0].size();
        if (((m + n - 1) & 1) || grid[0][0] == ')' || grid[m - 1][n - 1] == '(') return false;
        bool vis[m][n][(m + n + 1) / 2];
        memset(vis, 0, sizeof(vis));
        function<bool(int, int, int)> dfs = [&](int x, int y, int c) {
            if (c > m - x + n - y + 1) return false; // pruning, too many '('.
            if (x == m - 1 && y == n - 1) return c == 1;
            if (vis[x][y][c]) return false;
            vis[x][y][c] = true;
            c += grid[x][y] == '(' ? 1 : -1;
            return c >= 0 && (x < m - 1 && dfs(x + 1, y, c) || y < n - 1 && dfs(x, y + 1, c));
        };
        return dfs(0, 0, 0);
    }
};