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.

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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”.

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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

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Solution3

dynamic programming by using three dimension array

Get idea from here.