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 int n = s1.length(); if(n ==1){ return s1.equals(s2); } for(int i =1 ; i<=s2.length()/2 ; i++){ String x1 = s1.substring(0,i) , y1 = s1.substring(n-i), x2 = s2.substring(0,i), y2 = s2.substring(n-i); String a1 = s1.substring(i), b1 = s1.substring(0,n-i), a2 = s2.substring(i), b2 = s2.substring(0,n-i); boolean r1 = isScramble(x1,x2)&&isScramble(a1,a2); if(r1) return true; boolean r2 = isScramble(x1,y2)&&isScramble(a1,b2); if(r2) return true; boolean r3 = isScramble(y1,x2)&&isScramble(b1,a2); if(r3) return true; boolean r4 = isScramble(y1,y2)&&isScramble(b1,b2); if(r4) return true; } return false; } }