Android unlock patterns [Backtracking]¶
Time: O(92x29); Space: O(9x2^9); medium
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 1:
Input: m = 1, n = 1
Output: 9
Example 2:
Input: m = 1, n = 2
Output: 65
1. Dynamic Programming¶
[1]:
class Solution1(object):
"""
Time: O(9^2 * 2^9)
Space: O(9 * 2^9)
"""
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 range(1 << 9)]
for i in range(9):
dp[merge(0, i)][i] = 1
res = 0
for used in range(len(dp)):
number = number_of_keys(used)
if number > n:
continue
for i in range(9):
if not contain(used, i):
continue
if m <= number <= n:
res += dp[used][i]
x1, y1 = divmod(i, 3)
for j in range(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
[2]:
s = Solution1()
m = 1
n = 1
assert s.numberOfPatterns(m, n) == 9
m = 1
n = 2
assert s.numberOfPatterns(m, n) == 65
2. Dynamic Programming¶
[3]:
class Solution2(object):
"""
Time: O(9^2 * 2^9)
Space: O(9 * 2^9)
"""
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 range(1 << 9)]
for i in range(9):
dp[merge(0, i)][i] = 1
res = 0
for used in range(len(dp)):
number = number_of_keys(used)
if number > n:
continue
for i in range(9):
if not contain(used, i):
continue
x1, y1 = divmod(i, 3)
for j in range(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
[4]:
s = Solution2()
m = 1
n = 1
assert s.numberOfPatterns(m, n) == 9
m = 1
n = 2
assert s.numberOfPatterns(m, n) == 65
3. Backtracking solution (TLE)¶
[5]:
class Solution_TLE(object):
"""
Time: O(9!)
Space: O(9)
"""
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 range(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
[6]:
s = Solution_TLE()
m = 1
n = 1
assert s.numberOfPatterns(m, n) == 9
m = 1
n = 2
assert s.numberOfPatterns(m, n) == 65