dfs/android-unlock-patterns

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2023-12-01

Android Unlock Patterns

描述

Android devices have a special lock screen with a 3 x 3 grid of dots. Users can set an "unlock pattern" by connecting the dots in a specific sequence, forming a series of joined line segments where each segment's endpoints are two consecutive dots in the sequence. A sequence of k dots is a valid unlock pattern if both of the following are true:

All the dots in the sequence are distinct.
If the line segment connecting two consecutive dots in the sequence passes through any other dot, the other dot must have previously appeared in the sequence. No jumps through non-selected dots are allowed.

Here are some example valid and invalid unlock patterns:

The 1st pattern [4,1,3,6] is invalid because the line connecting dots 1 and 3 pass through dot 2, but dot 2 did not previously appear in the sequence.
The 2nd pattern [4,1,9,2] is invalid because the line connecting dots 1 and 9 pass through dot 5, but dot 5 did not previously appear in the sequence.
The 3rd pattern [2,4,1,3,6] is valid because it follows the conditions. The line connecting dots 1 and 3 meets the condition because dot 2 previously appeared in the sequence.
The 4th pattern [6,5,4,1,9,2] is valid because it follows the conditions. The line connecting dots 1 and 9 meets the condition because dot 5 previously appeared in the sequence.

Given two integers m and n, return the number of unique and valid unlock patterns of the Android grid lock screen that consist of at least m keys and at most n keys.

Two unlock patterns are considered unique if there is a dot in one sequence that is not in the other, or the order of the dots is different.

Example 1:

Input: m = 1, n = 1
Output: 9

Example 2:

Input: m = 1, n = 2
Output: 65

Constraints:

1 <= m, n <= 9

分析

深搜，配合大量剪枝。

剪枝通常需要深入理解题意，掌握领域知识（domain knowledge）。本题有个比较重要的特性是，1,3,5,7 这 4 个点是互相对称的，即以 1 作为起点，得到的有效模式的个数，与以 3 作为起点，得到的有效模式的个数，是相等的，5, 7 也是类似。同理，2,4,6,8 这 4 个点也是对称的。

换句话说，令 f[i]表示以i为起点的有效模式的个数，总数就是 f[1]*4+f[2]*4+f[5]。

代码

// Android Unlock Patterns
// Time Complexity O(n!), Space Complexity: O(n)
public class Solution {

    public int numberOfPatterns(int m, int n) {
        int[][] jumps = build_jump_table();
        boolean[] visited = new boolean[10];

        int count = 0;
        for(int i = m; i <= n; ++i) {
            count += dfs(visited, jumps, 1, i - 1) * 4;  // 1, 3, 7, 9 are symmetric
            count += dfs(visited, jumps, 2, i - 1) * 4;  // 2, 4, 6, 8 are symmetric
            count += dfs(visited, jumps, 5, i - 1);      // 5
        }
        return count;
    }

    private static int dfs(boolean visited[], int[][] jumps, int num, int remain) {
        if(remain == 0) return 1;

        int count = 0;
        visited[num] = true;
        for(int i = 1; i <= 9; ++i) {
            int mid = jumps[num][i]; // Edge num->i jumps over mid
            if(!visited[i] && (mid == 0 || (visited[mid]))) {
                count += dfs(visited, jumps, i, remain - 1);
            }
        }
        visited[num] = false;
        return count;
    }

    private int[][] build_jump_table() {
        int[][] jumps = new int[10][10]; // jump table, 0 means adjacent
        jumps[1][3] = jumps[3][1] = 2; // The edge 1->3 jumps over 2
        jumps[1][7] = jumps[7][1] = 4;
        jumps[3][9] = jumps[9][3] = 6;
        jumps[7][9] = jumps[9][7] = 8;
        jumps[2][8] = jumps[8][2] = jumps[4][6] = jumps[6][4] = 5;
        jumps[1][9] = jumps[9][1] = jumps[3][7] = jumps[7][3] = 5;
        return jumps;
    }
}

// TODO