Scramble String Leetcode Python

Given A string S1, we may represent it as a binary tree by partitioning it to the Non-empty substrings Recursivel Y.

Below is one possible representation of S1 = "great" :

    Great   /      gr    eat/\    /  g   r  E   at           /           a   t

To scramble the string, we are choose any non-leaf node and swap it to the children.

for example, if we choose the Node " GR "  and swap its-children, it produces a scrambled string < Code style= "Font-family:menlo,monaco,consolas, ' Courier New ', monospace; font-size:13px; PADDING:2PX 4px; Color:rgb (199,37,78); Background-color:rgb (249,242,244) ">" Rgeat ".

    Rgeat   /      RG    eat/\    /  r   G  e   at           /           a   t

We say is "rgeat" 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 is "rgtae" a scrambled string of "great" .

Given strings S1 and S2 of the same length, determine if S2 is a scrambled string of S1 .

This problem can is solved with brute force.

Class solution:    # @param s1, a string    # @param s2, a string    # @return A Boolean    def isscramble (self, S1, s 2):        N, m = Len (S1), Len (S2)        if n!=m or sorted (S1)!=sorted (S2):            return False        if n < 4 or S1 = = S2:            re Turn True for        I in range (1, N):            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])):                return True        return False                    

Scramble String Leetcode Python