package leetcode_1To300;

/**
 * 本代码来自 Cspiration，由 @Cspiration 提供
 * 题目来源：http://leetcode.com
 * - Cspiration 致力于在 CS 领域内帮助中国人找到工作，让更多海外国人受益
 * - 现有课程：Leetcode Java 版本视频讲解（1-900题）（上）（中）（下）三部
 * - 算法基础知识（上）（下）两部；题型技巧讲解（上）（下）两部
 * - 节省刷题时间，效率提高2-3倍，初学者轻松一天10题，入门者轻松一天20题
 * - 讲师：Edward Shi
 * - 官方网站：https://cspiration.com
 * - 版权所有，转发请注明出处
 */
public class _351_AndroidUnlockPatterns {
    /**
     * 351. Android Unlock Patterns
     * time : 不知道
     * space : 不知道
     * @param m
     * @param n
     * @return
     */

    public int numberOfPatterns(int m, int n) {
        int[][] skip = new int[10][10];
        skip[1][3] = skip[3][1] = 2;
        skip[1][7] = skip[7][1] = 4;
        skip[3][9] = skip[9][3] = 6;
        skip[7][9] = skip[9][7] = 8;
        skip[1][9] = skip[9][1] = skip[2][8] = skip[8][2] = skip[3][7] = skip[7][3] = skip[4][6] = skip[6][4] = 5;
        boolean[] visited = new boolean[10];
        int res = 0;
        for (int i = m; i <= n; i++) {
            res += DFS(visited, skip, 1, i - 1) * 4; // 1, 3, 7, 9
            res += DFS(visited, skip, 2, i - 1) * 4; // 2, 4, 6, 8
            res += DFS(visited, skip, 5, i - 1); // 5
        }
        return res;
    }

    public int DFS(boolean[] visited, int[][] skip, int cur, int remain) {
        if (remain < 0) return 0;
        if (remain == 0) return 1;
        visited[cur] = true;
        int res = 0;
        for (int i = 1; i <= 9; i++) {
            if (!visited[i] && (skip[cur][i] == 0 || visited[skip[cur][i]])) {
                res += DFS(visited, skip, i, remain - 1);
            }
        }
        visited[cur] = false;
        return res;
    }
}