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package com.fishercoder.solutions;
/**
* 87. 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 _87 {
/**credit: https://discuss.leetcode.com/topic/19158/accepted-java-solution*/
public boolean isScramble(String s1, String s2) {
if (s1.equals(s2)) return true;
int[] letters = new int[26];
for (int i = 0; i < s1.length(); i++) {
letters[s1.charAt(i) - 'a']++;
letters[s2.charAt(i) - 'a']--;
}
for (int i : letters) {
if (i != 0) return false;
}
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;
}
}