-
Notifications
You must be signed in to change notification settings - Fork 613
/
87.py
50 lines (41 loc) · 1.23 KB
/
87.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
'''
Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.
Below is one possible representation of s1 = "great":
great
/ \
gr eat
/ \ / \
g r e at
/ \
a t
To scramble the string, we may choose any non-leaf node and swap its two children.
For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".
rgeat
/ \
rg eat
/ \ / \
r g e at
/ \
a t
We say that "rgeat" is a scrambled string of "great".
'''
class Solution(object):
def __init__(self):
self.cache = {}
def isScramble(self, s1, s2):
if s1 == s2:
return True
if s1+s2 in self.cache:
return self.cache[s1+s2]
if len(s1) != len(s2) or sorted(s1) != sorted(s2):
self.cache[s1+s2] = False
return False
for index in range(1, len(s1)):
if self.isScramble(s1[:index], s2[:index]) and self.isScramble(s1[index:], s2[index:]):
self.cache[s1+s2] =True
return True
if self.isScramble(s1[:index], s2[-index:]) and self.isScramble(s1[index:], s2[0:-index]):
self.cache[s1+s2] = True
return True
self.cache[s1+s2] =False
return False