Leetcode 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.

Solution 1: Use DP, dp[i][j][k] means s1[i...i+k] & s2[j...j+k] is scramble string or not. It need 3 loop to iterate the dp array and another loop to calculate the dp[i][j][k], so complexity is O(n^4).

 public class Solution {  
   public boolean isScramble(String s1, String s2) {  
     int n=s1.length();  
     if (n!=s2.length()) return false;  
     boolean[][][] dp=new boolean[n][n][n];  
     for (int i=n-1; i>=0; i--) {  
       for (int j=n-1; j>=0; j--) {  
         for (int k=0; i+k<n && j+k<n; k++) {  
           if (k==0) dp[i][j][k]=s1.charAt(i)==s2.charAt(j);  
           else {  
             dp[i][j][k]=false;  
             for (int t=0; t<k; t++) {  
               if ((dp[i][j][t] && dp[i+t+1][j+t+1][k-t-1]) || (dp[i][j+k-t][t] && dp[i+t+1][j][k-t-1])) {  
                 dp[i][j][k]=true;  
                 break;  
               }  
             }  
           }  
         }  
       }  
     }  
     return dp[0][0][n-1];  
   }  
 }