LeetCode: Scramble String

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.

public class Solution {
    public boolean isScramble(String s1, String s2) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(s1==null || s2==null || s1.length()!=s2.length()) return false;
        char[]c1=s1.toCharArray(), c2=s2.toCharArray();
        Arrays.sort(c1);
        Arrays.sort(c2);
        if(!(new String(c1)).equals(new String(c2))) return false;
        else if(s1.length()==1)return true;
    
        for(int i=0;i<s1.length()-1;i++){
            if(isScramble(s1.substring(0,i+1),s2.substring(s1.length()-i-1))
                  && isScramble(s1.substring(i+1),s2.substring(0,s1.length()-i-1))
                  || isScramble(s1.substring(0,i+1),s2.substring(0,i+1)) 
                  && isScramble(s1.substring(i+1),s2.substring(i+1)))return true;
        }
        return false;
    }
}

import java.util.Map;
import java.util.HashMap;
import java.util.Stack;
import java.util.List;
import java.util.ArrayList;

public class Solution {
    public boolean isScramble(String s1, String s2) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(s1==null || s2==null || s1.length()!=s2.length()) return false;
        Map<Character,List<Integer>> order = new HashMap<Character,List<Integer>>(); 
        for(int i=0;i<s1.length();i++){
            if(order.containsKey(s1.charAt(i))){
                order.get(s1.charAt(i)).add(i);
            }else{
                List<Integer> l = new ArrayList<Integer>();
                l.add(i);
                order.put(s1.charAt(i),l);
            }
        }
        
        int[] indices = new int[s2.length()];
        for(int i=0;i<s2.length();i++){
            if(order.containsKey(s2.charAt(i))){
                indices[i]=order.get(s2.charAt(i)).get(0);
                order.get(s2.charAt(i)).remove(0);
                if(order.get(s2.charAt(i)).size()==0) 
                    order.remove(s2.charAt(i));
            }
            else return false;
        }
        
        Stack<List<Integer>> neighbors = new Stack<List<Integer>>();
        
        for(int i=0;i<indices.length;i++){
            List<Integer> cur = new ArrayList<Integer>();
            cur.add(indices[i]);
            cur.add(indices[i]);
            
            while(!neighbors.isEmpty()){
                List<Integer> temp = neighbors.peek();
                if(cur.get(1)==temp.get(0)-1){
                    temp.set(0,cur.get(0));
                }else if(cur.get(0)==temp.get(1)+1){
                        temp.set(1,cur.get(1));
                }else break;
                cur=neighbors.pop();
            }    
            neighbors.push(cur);
        }
        neighbors.pop();
        if(neighbors.isEmpty())return true;
        return false;
    }
}