87. Scramble String

Description

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":

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".

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".

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.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

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

Output: true

Example 2

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

Output: false

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1.length() != s2.length()) return false;
        
        if(s1.isEmpty() || s2.equals(s1)) return true;
        
        int len = s1.length();
        
        int[] countChars = new int[26];
        for(int i = 0; i < len; i++) {
            countChars[s1.charAt(i)-'a']++;
            countChars[s2.charAt(i)-'a']--;
        }
        for(int i = 0; i < 26; i++) {
            if(countChars[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;
    }
}

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