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.

Analysis

• Define scramble base case:
  • s1==s2
  • s1 and s2 has same chars

• "rgeat" and "great":
  • "rg" and "gr" is scramble (case 1)
  • "eat" and "eat" is scramble (case 1)

• "rgtae" and "great"
  • "rg" and "gr" is scramble (case 1)
  • "tae" and "eat":
    • "ta" and "at" is scramble (case 2)
    • "e" and "e" is scramble (case 2)

Code

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if (s1.equals(s2)) return true;

        int len = s1.length();
        int[] count= new int[26];
        for (int i=0; i<len; i++){
            count[s1.charAt(i)-'a']++;
            count[s2.charAt(i)-'a']--;
        }
        for (int i=0; i<26; i++){
            if (count[i]!=0) return false;
        }

        for (int i=1; i<len; i++){
            //case 1
            if (isScramble(s1.substring(0,i),s2.substring(0,i)) && isScramble(s1.substring(i),s2.substring(i))) return true;
            //case 2
            if (isScramble(s1.substring(0,i),s2.substring(len-i)) && isScramble(s1.substring(i),s2.substring(0,len-i))) return true;
        }

        return false;
    }
}

Reference

Leetcode