leetcode

Scramble String

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:

  public boolean isScramble(String s1, String s2) {      if(s1.equals(s2)) return true;      int len = s1.length();      int[] letters = new int[26];      for(int i=0;i<len;i++){          letters[s1.charAt(i) - 'a']++;          letters[s2.charAt(i) - 'a']--;      }      for(int i=0;i<26;i++){          if(letters[i] != 0) return false;      }      for(int i=1;i<len;i++){          if(isScramble(s1.substring(0,i),s2.substring(0,i)) && isScramble(s1.substring(i),s2.substring(i))) return true;          if(isScramble(s1.substring(0,i),s2.substring(len-i)) && isScramble(s1.substring(i),s2.substring(0,len-i))) return true;      }      return false;  }