Scramble String

Hard string

LeetCode: https://leetcode.com/problems/scramble-string/description/

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.

Solution

The 1st IF is to check the LEFT child of S1 is scramble of LEFT child of S2 AND RIGHT child of S1 is also a scramble of RIGHT child of s2.
When this fails, it means the left and right substrings are swapped.
The 2nd IF statement check for the swap case with LEFT child of S1 and RIGHT child of S2 AND RIGHT child of S1 and LEFT child of S2.

see: https://discuss.leetcode.com/topic/19158/accepted-java-solution/8

Java

public class Solution {
    public boolean isScramble(String s1, String s2) {
        // using recurision to check substring
        if (s1.equals(s2)) {
            return true;
        }
        if (s1.length() != s2.length()) {
            return false;
        }
        int[] counts = new int[256];
        for (int i = 0; i < s1.length(); i++) {
            ++counts[s1.charAt(i)];
            --counts[s2.charAt(i)];
        }
        for (int i = 0; i < s1.length(); i++) {
            if (counts[s1.charAt(i)] != 0) {
                return false;
            }
        }
        for (int i = 1; i < s1.length(); i++) { // i starts with 1, not 0
            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;
    }
}