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Leetcode087-scrambleString

Posted on Edited on In Views: Valine:
Solution Report of LeetCode Acceptted

Description

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”:

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    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”.

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    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”.

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    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

Example 1:

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Input: s1 = "great", s2 = "rgeat"
Output: true

Example 2:
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Input: s1 = "abcde", s2 = "caebd"
Output: false

Solution

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