Some Interesting Solution - Android Unlock Patterns


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    Android Unlock Patterns

    public class Solution {
        // cur: the current position
        // remain: the steps remaining
        int dfs(boolean vis[], int[][] skip, int cur, int remain) {
            if(remain < 0) return 0;
            if(remain == 0) return 1;
            vis[cur] = true;
            int rst = 0;
            for(int i = 1; i <= 9; ++i) {
                // If vis[i] is not visited and (two numbers are adjacent or skip number is already visited)
                if(!vis[i] && (skip[cur][i] == 0 || (vis[skip[cur][i]]))) {
                    rst += dfs(vis, skip, i, remain - 1);
                }
            }
            vis[cur] = false;
            return rst;
        }
        
        public int numberOfPatterns(int m, int n) {
            // Skip array represents number to skip between two pairs
            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 vis[] = new boolean[10];
            int rst = 0;
            // dfs search each length from m to n
            for(int i = m; i <= n; ++i) {
                rst += dfs(vis, skip, 1, i - 1) * 4;    // 1, 3, 7, 9 are symmetric
                rst += dfs(vis, skip, 2, i - 1) * 4;    // 2, 4, 6, 8 are symmetric
                rst += dfs(vis, skip, 5, i - 1);        // 5
            }
            return rst;
        }
    };
    

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