Given an Android 3x3 key lock screen and two integers m and n, where 1 ≤ m ≤ n ≤ 9, count the total number of unlock patterns of the Android lock screen, which consist of minimum of m keys and maximum n keys.
Rules for a valid pattern:
- Each pattern must connect at least m keys and at most n keys.
- All the keys must be distinct.
- 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.
- The order of keys used matters.
Explanation:
| 1 | 2 | 3 | | 4 | 5 | 6 | | 7 | 8 | 9 |
Invalid move: 4 - 1 - 3 - 6
Line 1 - 3 passes through key 2 which had not been selected in the pattern.
Invalid move: 4 - 1 - 9 - 2
Line 1 - 9 passes through key 5 which had not been selected in the pattern.
Valid move: 2 - 4 - 1 - 3 - 6
Line 1 - 3 is valid because it passes through key 2, which had been selected in the pattern
Valid move: 6 - 5 - 4 - 1 - 9 - 2
Line 1 - 9 is valid because it passes through key 5, which had been selected in the pattern.
Example:
Given m = 1, n = 1, return 9.
在安卓的3*3的解锁屏幕上，给出2个整数m, n(1 ≤ m ≤ n ≤ 9)，问在m到n的滑动次数之间，有多少种可能的解锁方案。给出了合理和不合理的滑动。
优化方法是，由于 1,3,7,9 是对称的，2,4,6,8也是对称的，所以只用计算其中一个，然后乘以4，5是单独的一个，所以总共求3组就可以了。
解法：DFS，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，然后用递归来解，先对1调用递归函数，在递归函数中遍历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，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。参考
Java:
public class Solution {
private int patterns;
private boolean valid(boolean[] keypad, int from, int to) {
if (from==to) return false;
int i=Math.min(from, to), j=Math.max(from,to);
if ((i==1 && j==9) || (i==3 && j==7)) return keypad[5] && !keypad[to];
if ((i==1 || i==4 || i==7) && i+2==j) return keypad[i+1] && !keypad[to];
if (i<=3 && i+6==j) return keypad[i+3] && !keypad[to];
return !keypad[to];
}
private void find(boolean[] keypad, int from, int step, int m, int n) {
if (step == n) {
patterns ++;
return;
}
if (step >= m) patterns ++;
for(int i=1; i<=9; i++) {
if (valid(keypad, from, i)) {
keypad[i] = true;
find(keypad, i, step+1, m, n);
keypad[i] = false;
}
}
}
public int numberOfPatterns(int m, int n) {
boolean[] keypad = new boolean[10];
for(int i=1; i<=9; i++) {
keypad[i] = true;
find(keypad, i, 1, m, n);
keypad[i] = false;
}
return patterns;
}
}
Java:
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;
}
}
Python:
# Time: O(9!)
# Space: O(9)
# Backtracking solution. (TLE)
class Solution_TLE(object):
def numberOfPatterns(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
def merge(used, i):
return used | (1 << i)
def contain(used, i):
return bool(used & (1 << i))
def convert(i, j):
return 3 * i + j
def numberOfPatternsHelper(m, n, level, used, i):
number = 0
if level > n:
return number
if m <= level <= n:
number += 1
x1, y1 = divmod(i, 3)
for j in xrange(9):
if contain(used, j):
continue
x2, y2 = divmod(j, 3)
if ((x1 == x2 and abs(y1 - y2) == 2) or
(y1 == y2 and abs(x1 - x2) == 2) or
(abs(x1 - x2) == 2 and abs(y1 - y2) == 2)) and
not contain(used,
convert((x1 + x2) // 2, (y1 + y2) // 2)):
continue
number += numberOfPatternsHelper(m, n, level + 1, merge(used, j), j)
return number
number = 0
# 1, 3, 7, 9
number += 4 * numberOfPatternsHelper(m, n, 1, merge(0, 0), 0)
# 2, 4, 6, 8
number += 4 * numberOfPatternsHelper(m, n, 1, merge(0, 1), 1)
# 5
number += numberOfPatternsHelper(m, n, 1, merge(0, 4), 4)
return number
Python:
# Time: O(9^2 * 2^9)
# Space: O(9 * 2^9)
# DP solution.
class Solution2(object):
def numberOfPatterns(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
def merge(used, i):
return used | (1 << i)
def number_of_keys(i):
number = 0
while i > 0:
i &= i - 1
number += 1
return number
def exclude(used, i):
return used & ~(1 << i)
def contain(used, i):
return bool(used & (1 << i))
def convert(i, j):
return 3 * i + j
# dp[i][j]: i is the set of the numbers in binary representation,
# d[i][j] is the number of ways ending with the number j.
dp = [[0] * 9 for _ in xrange(1 << 9)]
for i in xrange(9):
dp[merge(0, i)][i] = 1
res = 0
for used in xrange(len(dp)):
number = number_of_keys(used)
if number > n:
continue
for i in xrange(9):
if not contain(used, i):
continue
x1, y1 = divmod(i, 3)
for j in xrange(9):
if i == j or not contain(used, j):
continue
x2, y2 = divmod(j, 3)
if ((x1 == x2 and abs(y1 - y2) == 2) or
(y1 == y2 and abs(x1 - x2) == 2) or
(abs(x1 - x2) == 2 and abs(y1 - y2) == 2)) and
not contain(used,
convert((x1 + x2) // 2, (y1 + y2) // 2)):
continue
dp[used][i] += dp[exclude(used, i)][j]
if m <= number <= n:
res += dp[used][i]
return res
Python:
# DP solution.
class Solution(object):
def numberOfPatterns(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
def merge(used, i):
return used | (1 << i)
def number_of_keys(i):
number = 0
while i > 0:
i &= i - 1
number += 1
return number
def contain(used, i):
return bool(used & (1 << i))
def convert(i, j):
return 3 * i + j
# dp[i][j]: i is the set of the numbers in binary representation,
# dp[i][j] is the number of ways ending with the number j.
dp = [[0] * 9 for _ in xrange(1 << 9)]
for i in xrange(9):
dp[merge(0, i)][i] = 1
res = 0
for used in xrange(len(dp)):
number = number_of_keys(used)
if number > n:
continue
for i in xrange(9):
if not contain(used, i):
continue
if m <= number <= n:
res += dp[used][i]
x1, y1 = divmod(i, 3)
for j in xrange(9):
if contain(used, j):
continue
x2, y2 = divmod(j, 3)
if ((x1 == x2 and abs(y1 - y2) == 2) or
(y1 == y2 and abs(x1 - x2) == 2) or
(abs(x1 - x2) == 2 and abs(y1 - y2) == 2)) and
not contain(used,
convert((x1 + x2) // 2, (y1 + y2) // 2)):
continue
dp[merge(used, j)][j] += dp[used][i]
return res
C++:
// DP solution.
class Solution {
public:
int numberOfPatterns(int m, int n) {
// dp[i][j]: i is the set of the numbers in binary representation,
// dp[i][j] is the number of ways ending with the number j.
vector<vector<int>> dp(1 << 9 , vector<int>(9, 0));
for (int i = 0; i < 9; ++i) {
dp[merge(0, i)][i] = 1;
}
int res = 0;
for (int used = 0; used < dp.size(); ++used) {
const auto number = number_of_keys(used);
if (number > n) {
continue;
}
for (int i = 0; i < 9; ++i) {
if (!contain(used, i)) {
continue;
}
if (m <= number && number <= n) {
res += dp[used][i];
}
const auto x1 = i / 3;
const auto y1 = i % 3;
for (int j = 0; j < 9; ++j) {
if (contain(used, j)) {
continue;
}
const auto x2 = j / 3;
const auto y2 = j % 3;
if (((x1 == x2 && abs(y1 - y2) == 2) ||
(y1 == y2 && abs(x1 - x2) == 2) ||
(abs(x1 - x2) == 2 && abs(y1 - y2) == 2)) &&
!contain(used, convert((x1 + x2) / 2, (y1 + y2) / 2))) {
continue;
}
dp[merge(used, j)][j] += dp[used][i];
}
}
}
return res;
}
private:
inline int merge(int i, int j) {
return i | (1 << j);
}
inline int number_of_keys(int i) {
int number = 0;
for (; i; i &= i - 1) {
++number;
}
return number;
}
inline bool contain(int i, int j) {
return i & (1 << j);
}
inline int convert(int i, int j) {
return 3 * i + j;
}
};
C++:
// Time: O(9^2 * 2^9)
// Space: O(9 * 2^9)
// DP solution.
class Solution2 {
public:
int numberOfPatterns(int m, int n) {
// dp[i][j]: i is the set of the numbers in binary representation,
// dp[i][j] is the number of ways ending with the number j.
vector<vector<int>> dp(1 << 9 , vector<int>(9, 0));
for (int i = 0; i < 9; ++i) {
dp[merge(0, i)][i] = 1;
}
int res = 0;
for (int used = 0; used < dp.size(); ++used) {
const auto number = number_of_keys(used);
if (number > n) {
continue;
}
for (int i = 0; i < 9; ++i) {
if (!contain(used, i)) {
continue;
}
const auto x1 = i / 3;
const auto y1 = i % 3;
for (int j = 0; j < 9; ++j) {
if (i == j || !contain(used, j)) {
continue;
}
const auto x2 = j / 3;
const auto y2 = j % 3;
if (((x1 == x2 && abs(y1 - y2) == 2) ||
(y1 == y2 && abs(x1 - x2) == 2) ||
(abs(x1 - x2) == 2 && abs(y1 - y2) == 2)) &&
!contain(used, convert((x1 + x2) / 2, (y1 + y2) / 2))) {
continue;
}
dp[used][i] += dp[exclude(used, i)][j];
}
if (m <= number && number <= n) {
res += dp[used][i];
}
}
}
return res;
}
private:
inline int merge(int i, int j) {
return i | (1 << j);
}
inline int number_of_keys(int i) {
int number = 0;
for (; i; i &= i - 1) {
++number;
}
return number;
}
inline bool contain(int i, int j) {
return i & (1 << j);
}
inline int exclude(int i, int j) {
return i & ~(1 << j);
}
inline int convert(int i, int j) {
return 3 * i + j;
}
};
C++:
// Time: O(9!)
// Space: O(9)
// Backtracking solution.
class Solution3 {
public:
int numberOfPatterns(int m, int n) {
int number = 0;
// 1, 3, 5, 7
number += 4 * numberOfPatternsHelper(m, n, 1, merge(0, 0), 0);
// 2, 4, 6, 8
number += 4 * numberOfPatternsHelper(m, n, 1, merge(0, 1), 1);
// 5
number += numberOfPatternsHelper(m, n, 1, merge(0, 4), 4);
return number;
}
private:
int numberOfPatternsHelper(int m, int n, int level, int used, int i) {
int number = 0;
if (level > n) {
return number;
}
if (level >= m) {
++number;
}
const auto x1 = i / 3;
const auto y1 = i % 3;
for (int j = 0; j < 9; ++j) {
if (contain(used, j)) {
continue;
}
const auto x2 = j / 3;
const auto y2 = j % 3;
if (((x1 == x2 && abs(y1 - y2) == 2) ||
(y1 == y2 && abs(x1 - x2) == 2) ||
(abs(x1 - x2) == 2 && abs(y1 - y2) == 2)) &&
!contain(used, convert((x1 + x2) / 2, (y1 + y2) / 2))) {
continue;
}
number += numberOfPatternsHelper(m, n, level + 1, merge(used, j), j);
}
return number;
}
private:
inline int merge(int i, int j) {
return i | (1 << j);
}
inline bool contain(int i, int j) {
return i & (1 << j);
}
inline int convert(int i, int j) {
return 3 * i + j;
}
};
C++:
class Solution {
public:
int DFS(int m, int n, int len, int num)
{
int cnt = 0;
if(len >= m) cnt++;
if(++len > n) return cnt;
visited[num] = true;
for(int i = 1; i<= 9; i++)
if(!visited[i] && visited[hash[num][i]])
cnt += DFS(m, n, len, i);
visited[num] = false;
return cnt;
}
int numberOfPatterns(int m, int n) {
if(m < 1 || n < 1) return 0;
visited.resize(10, false);
visited[0] = true;
hash.resize(10, vector<int>(10, 0));
hash[1][3] = hash[3][1] = 2;
hash[1][7] = hash[7][1] = 4;
hash[3][9] = hash[9][3] = 6;
hash[7][9] = hash[9][7] = 8;
hash[2][8] = hash[8][2] = hash[4][6] = hash[6][4] = 5;
hash[1][9] = hash[9][1] = hash[3][7] = hash[7][3] = 5;
return DFS(m, n, 1, 1)*4 + DFS(m, n, 1, 2)*4 + DFS(m, n, 1, 5);
}
private:
vector<bool> visited;
vector<vector<int>> hash;
};
C++:
class Solution {
public:
int numberOfPatterns(int m, int n) {
return count(m, n, 0, 1, 1);
}
int count(int m, int n, int used, int i1, int j1) {
if (n == 0) return 1;
int res = (m <= 0);
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
// used2 check middle point has been used
int I = i1+i, J = j1+j, used2 = used | (1 << (i*3+j));
// used2 > used: add a new unused integer
// I%2 == 1: i1 odd i even or reverse
// used2 & (1 << I/2*3+J/2): mid point has been used
if (used2 > used && (I%2 || J%2 || used2 & (1 << I/2*3+J/2))) {
res += count(m-1, n-1, used2, i, j);
}
}
}
return res;
}
};