Scramble_String.py

"""
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.
"""

class Solution:
    # @return a boolean
    def isScramble(self, s1, s2):
        if len(s1) != len(s2):
            return False
        if not self.hasSameLetter(s1, s2):
            return False
        if len(s1) <= 2:
            return True
        for i in range(1, len(s1)):
            if ( self.isScramble(s1[:i], s2[:i]) and self.isScramble(s1[i:], s2[i:]) ) or ( self.isScramble(s1[:i], s2[-i:]) and self.isScramble(s1[i:], s2[:-i]) ): # This is soooo important, -i!!!
                return True
        return False

    def hasSameLetter(self, s1, s2):
        if sorted(s1) != sorted(s2):
            return False
        return True
    # Another way to do this in dp, need to learn