Thursday, June 2, 2016

LeetCode 351 - Android Unlock Patterns


Related: Android patterns possible on 3x3 matrix of numbers
http://lcoj.tk/problems/android-unlock-patternselmirap/
Given Android 9 key lock screen and numbers m and n, where 1 <= m <= n <= 9 .
Count the total number of patterns of Android lock screen, which consist of minimum of m keys and maximum n keys.

Rules for valid pattern

  1. Each pattern must connect at least m keys and at most n keys
  2. All the keys must be distinct
  3. If the line connecting two consecutive keys in the pattern passes through any other keys, the other keys must have previously selected in the pattern. No jumps through non selected key is allowed
  4. The order of keys used matters.

Example:

| 1 | 2 | 3 |
| 4 | 5 | 6 |
| 7 | 8 | 9 |

Valid move : 6 - 5 - 4 - 1 - 9 - 2

Line 1 - 9 is valid because it pass through key 5, which has been already selected in the pattern

Valid move : 2 - 4 - 1 - 3 - 6

Line 1 - 3 is valid because it pass through key 2, which has been already selected in the pattern

Invalid move : 4 - 1 - 3 - 6

Line 1 - 3 pass through key 2 which is not still selected in the pattern

Invalid move : 4 - 1 - 9 - 2

Line 1 - 9 pass through key 5 which is not still selected in the pattern
https://leetcode.com/discuss/104500/java-solution-with-clear-explanations-and-optimization-81ms
The optimization idea is that 1,3,7,9 are symmetric, 2,4,6,8 are also symmetric. Hence we only calculate one among each group and multiply by 4.
// 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[i][cur] == 0 || (vis[skip[i][cur]]))) { 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; }
https://leetcode.com/discuss/104688/simple-and-concise-java-solution-in-69ms
The general idea is DFS all the possible combinations from 1 to 9 and skip invalid moves along the way.
We can check invalid moves by using a jumping table. e.g. If a move requires a jump and the key that it is crossing is not visited, then the move is invalid. Furthermore, we can utilize symmetry to reduce runtime, in this case it reduced from ~120ms to ~70ms.
https://leetcode.com/discuss/104311/java-easy-understand-dfs-solution-72ms
If we use the symmetry, we can only start from 1, 2 and 5 then multiply the results of 1 and 2 by 4. (170ms)
http://dartmooryao.blogspot.com/2016/05/leetcode-351-android-unlock-patterns.html
(1) Create an grid, list all condition paths. For example, if we are currently at node 1, and we want to go to node 3, then we need node 2 visited. We use condPaths[1][3] = 2; to represent this relationship.
(2) Use an boolean array to keep track of the nodes we have visited before.
(3) Count the path number from min to max seperately.
(4) Use recursion function, given current position, check all possible paths. If the next node is not visited yet, and it satisfies the constraint, then we can go to this path.
(5) Return 1 when we find the end of path.
    public int numberOfPatterns(int m, int n) {
        int[][] condPaths = getCondPath();
        boolean[] visited = new boolean[10];
        int totalCount = 0;
        for(int i=m; i<=n; i++){
            for(int j=1; j<=9; j++){
                totalCount += getPathCount(i, j, visited, condPaths);               
            }
        }
        return totalCount;
    }
  
    private int getPathCount(int pathN, int currPosi, boolean[] visited, int[][] condPaths){
        if(pathN == 1){ return 1; }
        int count = 0;
        visited[currPosi] = true;
        for(int i=1; i<visited.length; i++){
            if(!visited[i] && (condPaths[currPosi][i] == 0 || visited[condPaths[currPosi][i]])){
                count += getPathCount(pathN-1, i, visited, condPaths);
            }
        }
        visited[currPosi] = false;
        return count;
    }
  
    private int[][] getCondPath(){
        int[][] condPaths = new int[10][10];
        condPaths[1][7] = 4;
        condPaths[1][3] = 2;
        condPaths[1][9] = 5;
        condPaths[2][8] = 5;
        condPaths[3][1] = 2;
        condPaths[3][9] = 6;
        condPaths[3][7] = 5;
        condPaths[4][6] = 5;
        condPaths[6][4] = 5;
        condPaths[7][1] = 4;
        condPaths[7][9] = 8;
        condPaths[7][3] = 5;
        condPaths[8][2] = 5;
        condPaths[9][7] = 8;
        condPaths[9][3] = 6;
        condPaths[9][1] = 5;
        return condPaths;
    }
https://leetcode.com/discuss/104311/java-easy-understand-dfs-solution-72ms
If we use the symmetry, we can only start from 1, 2 and 5 then multiply the results of 1 and 2 by 4. (170ms)

http://www.cnblogs.com/grandyang/p/5541012.html
我们建立一个二维数组jumps,用来记录两个数字键之间是否有中间键,然后再用一个一位数组visited来记录某个键是否被访问过,然后我们用递归来解,我们先对1调用递归函数,在递归函数中,我们遍历1到9每个数字next,然后找他们之间是否有jump数字,如果next没被访问过,并且jump为0,或者jump被访问过,我们对next调用递归函数。数字1的模式个数算出来后,由于1,3,7,9是对称的,所以我们乘4即可,然后再对数字2调用递归函数,2,4,6,9也是对称的,再乘4,最后单独对5调用一次,然后把所有的加起来就是最终结果了
    int numberOfPatterns(int m, int n) {
        int res = 0;
        vector<bool> visited(10, false);
        vector<vector<int>> jumps(10, vector<int>(10, 0));
        jumps[1][3] = jumps[3][1] = 2;
        jumps[4][6] = jumps[6][4] = 5;
        jumps[7][9] = jumps[9][7] = 8;
        jumps[1][7] = jumps[7][1] = 4;
        jumps[2][8] = jumps[8][2] = 5;
        jumps[3][9] = jumps[9][3] = 6;
        jumps[1][9] = jumps[9][1] = jumps[3][7] = jumps[7][3] = 5;
        res += helper(1, 1, 0, m, n, jumps, visited) * 4;
        res += helper(2, 1, 0, m, n, jumps, visited) * 4;
        res += helper(5, 1, 0, m, n, jumps, visited);
        return res;
    }
    int helper(int num, int len, int res, int m, int n, vector<vector<int>> &jumps, vector<bool> &visited) {
        if (len >= m) ++res;
        ++len;
        if (len > n) return res;
        visited[num] = true;
        for (int next = 1; next <= 9; ++next) {
            int jump = jumps[num][next];
            if (!visited[next] && (jump == 0 || visited[jump])) {
                res = helper(next, len, res, m, n, jumps, visited);
            }
        }
        visited[num] = false;
        return res;
    }

X.
https://leetcode.com/discuss/104320/short-c-solution
其中used是一个9位的mask,每位对应一个数字,如果为1表示存在,0表示不存在,(i1, j1)是之前的位置,(i, j)是当前的位置,所以滑动是从(i1, j1)到(i, j),中间点为((i1+i)/2, (j1+j)/2), 这里的I和J分别为i1+i和j1+j,还没有除以2,所以I和J都是整数。如果I%2或者J%2不为0,说明中间点的坐标不是整数,即中间点不存在,如果中间点存在,如果中间点被使用了,则这条线也是成立的,可以调用递归

used is the 9-bit bitmask telling which keys have already been used and (i1,j1) and (i2,j2)are the previous two key coordinates. A step is valid if...
  • I % 2: It goes to a neighbor row or
  • J % 2: It goes to a neighbor column or
  • used2 & (1 << (I/2*3 + J/2))): The key in the middle of the step has already been used.
(i2,j2) are the coordinates of the previous key, (i,j) are the coordinates of the new key. So the new line goes from (i2,j2) to (i,j). The middle point of the line is at ((i2+i)/2, (j2+j)/2). My Iand J are those middle coordinates, except I didn't divide by 2 yet, so I can stay in integers.
Now if I % 2 isn't zero, then that means I/2 and thus (i2+i)/2 is no integer. Which means the middle point of the line is not a key. Same with the other coordinate. If both those checks fail, then the middle point of the new line is a key, and thus I need to check that that key has been used already.
int numberOfPatterns(int m, int n) { return count(m, n, 0, 1, 1, 1, 1); } private: int count(int m, int n, int used, int i1, int j1, int i2, int j2) { int number = m <= 0; if (!n) return 1; for (int i=0; i<3; i++) { for (int j=0; j<3; j++) { int I = i2 + i, J = j2 + j, used2 = used | (1 << (i*3 + j)); if (used2 > used && (I % 2 || J % 2 || used2 & (1 << (I/2*3 + J/2)))) number += count(m-1, n-1, used2, i2, j2, i, j); } } return number; }
http://www.geeksforgeeks.org/number-of-ways-to-make-mobile-lock-pattern/



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