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.
1 public class ScrambleString {
2 public boolean isScramble(String s1, String s2) {
3 if (s1.length() != s2.length())
4 return false;
5 if (s1.equals(s2))
6 return true;
7
8 int[] A = new int[26];
9 for (int i = 0; i < s1.length(); i++) {
10 ++A[s1.charAt(i) - 'a'];
11 }
12
13 for (int j = 0; j < s2.length(); j++) {
14 --A[s2.charAt(j) - 'a'];
15 }
16
17 for (int k = 0; k < 26; k++) {
18 if (A[k] != 0)
19 return false;
20 }
21
22 for (int i = 1; i < s1.length(); i++) {
23 boolean result = isScramble(s1.substring(0, i), s2.substring(0, i))
24 && isScramble(s1.substring(i), s2.substring(i));
25 result = result
26 || (isScramble(s1.substring(0, i),
27 s2.substring(s2.length() - i, s2.length())) && isScramble(
28 s1.substring(i), s2.substring(0, s2.length() - i)));
29 if (result)
30 return true;
31 }
32 return false;
33 }
34 }