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 {
/**
* @param s1 A string
* @param s2 Another string
* @return whether s2 is a scrambled string of s1
*/
public boolean isScramble(String s1, String s2) {
if (s1.length() != s2.length()) return false;

if (s1.equals(s2)) return true;

if (!isValid(s1, s2)) {
return false;
}

for (int i = 1; i < s1.length(); i++) {
String s11 = s1.substring(0, i);
String s12 = s1.substring(i);

String s21 = s2.substring(0, i);
String s22 = s2.substring(i);

String s23 = s2.substring(0, s2.length() - i);
String s24 = s2.substring(s2.length() - i);

if (isScramble(s11, s21) && isScramble(s12, s22) || isScramble(s11, s24) && isScramble(s12, s23)) {
return true;
}
}
return false;
}

private boolean isValid(String s1, String s2) {
char[] arr1 = s1.toCharArray();
char[] arr2 = s2.toCharArray();
Arrays.sort(arr1);
Arrays.sort(arr2);
if (!(new String(arr1)).equals(new String(arr2))) {
return false;
}
return true;
}
}