Scramble string

Time: O(N^4); Space: O(N^3); hard

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.

Example 1:

Input: s1 = “great”, s2 = “rgeat”

Output: True

Explanation:

  • As described above.

Example 2:

Input: s1 = “abcde”, s2 = “caebd”

Output: False

[1]:

class Solution1(object):
    """
    Time: O(N^4)
    Space: O(N^3)
    """
    def isScramble(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: bool
        """
        if not s1 or not s2 or len(s1) != len(s2):
            return False

        if s1 == s2:
            return True

        result = [[[False for j in range(len(s2))] for i in range(len(s1))] for n in range(len(s1) + 1)]

        for i in range(len(s1)):
            for j in range(len(s2)):
                if s1[i] == s2[j]:
                    result[1][i][j] = True

        for n in range(2, len(s1) + 1):
            for i in range(len(s1) - n + 1):
                for j in range(len(s2) - n + 1):
                    for k in range(1, n):
                        if result[k][i][j] and result[n - k][i + k][j + k] or\
                           result[k][i][j + n - k] and result[n - k][i + k][j]:
                            result[n][i][j] = True
                            break

        return result[n][0][0]

[2]:

s = Solution1()

s1 = "great"
s2 = "rgeat"
assert s.isScramble(s1, s2) == True

s1 = "abcde"
s2 = "caebd"
assert s.isScramble(s1, s2) == False