87. 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.

Example 1:

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

Example 2:

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

Here is a iterative DP solution https://leetcode.com/problems/scramble-string/discuss/29396/Simple-iterative-DP-Java-solution-with-explanation

for some 1 <= q < k we have: F(i, j, k) = (F(i, j, q) AND F(i + q, j + q, k - q)) OR (F(i, j + k - q, q) AND F(i + q, j, k - q))

  • Recursive
class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1.length() != s2.length()) return false;
        if(s1.equals(s2)) return true;
        if(s1.length() == 1) return false;
        int[] cnt = new int[26];
        for(char c : s1.toCharArray()) cnt[c-'a']++;
        for(char c : s2.toCharArray()) cnt[c-'a']--;
        for(int i : cnt) if(i != 0) return false;

        for(int i = 1; i < s1.length(); i++) {
            String s11 = s1.substring(0, i), s12 = s1.substring(i);
            String s21 = s2.substring(0, i), s22 = s2.substring(i);
            boolean r1 = isScramble(s11, s21) && isScramble(s12, s22);
            if(r1) return true;
            s21 = s2.substring(0, s1.length()-i);
            s22 = s2.substring(s1.length()-i);
            boolean r2 = isScramble(s11, s22) && isScramble(s12, s21);
            if(r2) return true;
        }
        return false;
    }
}