Leetcode: Scramble String
Question
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”.
Similarly, if we continue to swap the children of nodes “eat” and “at”, it produces a scrambled string “rgtae”.
rgtae
/ \
rg tae
/ \ / \
r g ta e
/ \
t a
We say that “rgtae” is a scrambled string of “great”.
Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.
Solution 1 (Wrong)
Code
class Solution(object):
def isScramble(self, s1, s2):
""" :type s1: str :type s2: str :rtype: bool """
s1,s2 = list(s1), list(s2)
return self.helper(s1, 0, len(s1)-1, s2, 0, len(s2)-1)
def helper(self, s1, a1, b1, s2, a2, b2):
if b1-a1 != b2-a2:
return False
if a1==b1:
return s1[a1]==s2[a2]
for ind in range(a1,b1):
# left-left, right-right
if self.equals(s1[a1:ind+1], s2[a2:a2+ind-a1+1]) and self.equals(s1[ind+1:b1+1], s2[b2-(b1-ind-1):b2+1]):
return True
# left-right, right-left
if self.equals(s1[a1:ind+1], s2[b2-(ind-a1):b2+1]) and self.equals(s1[ind+1:b1+1], s2[a2:a2+b1-ind]):
return True
return False
def equals(self, s1, s2):
""" to determine whether s1, s2 contain the same chars. """
dictin={}
for elem in s1:
if elem not in dictin.keys():
dictin[elem] = 1
else:
dictin[elem] += 1
for elem in s2:
if elem not in dictin.keys():
dictin[elem] = -1
else:
dictin[elem] -= 1
for value in dictin.values():
if value!=0:
return False
return True
Analysis
My fault is that even s1, s2 have the same chars, they are not scrambled. For instance, s1=”abcd”, s2=”bdac”.
Solution2
Analysis
Get idea from here.
Code
class Solution(object):
def isScramble(self, s1, s2):
""" :type s1: str :type s2: str :rtype: bool """
s1,s2 = list(s1), list(s2)
return self.helper(s1, 0, len(s1)-1, s2, 0, len(s2)-1)
def helper(self, s1, a1, b1, s2, a2, b2):
if b1-a1 != b2-a2:
return False
if a1==b1:
return s1[a1]==s2[a2]
if self.equals(s1[a1:b1+1], s2[a2:b2+1])==False:
return False
for ind in range(a1,b1):
# left-left, right-right
if self.helper(s1,a1,ind, s2,a2,a2+ind-a1) and self.helper(s1,ind+1,b1, s2,b2-(b1-ind-1),b2):
return True
# left-right, right-left
if self.helper(s1,a1,ind, s2,b2-(ind-a1),b2) and self.helper(s1,ind+1,b1, s2,a2,a2+b1-ind-1):
return True
return False
def equals(self, s1, s2):
""" to determine whether s1, s2 contain the same chars. """
dictin={}
for elem in s1:
if elem not in dictin.keys():
dictin[elem] = 1
else:
dictin[elem] += 1
for elem in s2:
if elem not in dictin.keys():
dictin[elem] = -1
else:
dictin[elem] -= 1
for value in dictin.values():
if value!=0:
return False
return True
Solution3
dynamic programming by using three dimension array
Get idea from here.