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android-unlock-patterns.py
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android-unlock-patterns.py
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# Time: O(9^2 * 2^9)
# Space: O(9 * 2^9)
# 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
# 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
# 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