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