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LeetCode-Solutions/Python/android-unlock-patterns.py at master · cppp/LeetCode-Solutions · GitHub
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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
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