leetcode - 87.Scramble String

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.

Solution:

public boolean isScramble(String s1, String s2) {
if(s1.equals(s2)) return true;

int len = s1.length();
int[] letters = new int[26];
for(int i=0;i<len;i++){
letters[s1.charAt(i) - 'a']++;
letters[s2.charAt(i) - 'a']--;
}
for(int i=0;i<26;i++){
if(letters[i] != 0) return false;
}

for(int i=1;i<len;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(len-i)) && isScramble(s1.substring(i),s2.substring(0,len-i))) return true;
}
return false;
}