# Question

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.

Example 1:

Input: s1 = "great", s2 = "rgeat"
Output: true


Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false


# Solution

Divide-and-Conquer.

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