LeetCode 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.

Solution 1:

public boolean isScramble(String s1, String s2) {
    if(s1.equals(s2)) return true; // base case
    if(s1.length() != s2.length()) return false; 
    int n = s1.length();
    char[] s1Array = s1.toCharArray();  
    char[] s2Array = s2.toCharArray();  
    Arrays.sort(s2Array);  
    Arrays.sort(s1Array);  
    for(int i = 0; i < n; i++){  
        if(s1Array[i] != s2Array[i])  
            return false;  
    }  
    
    for(int i = 1; i < n;i++){  
        String s1pre = s1.substring(0,i);  
        String s1suf = s1.substring(i);  
        String s2pre = s2.substring(0, i);  
        String s2suf = s2.substring(i);  
        boolean result = isScramble(s1pre, s2pre) && isScramble(s1suf, s2suf);  
        if(!result){  
            s2pre = s2.substring(0, s1.length() - i);  
            s2suf = s2.substring(s1.length() - i);  
            result = isScramble(s1pre, s2suf) && isScramble(s1suf, s2pre);  
        }  
        if(result) return true;  
    }  
    return false;  
}

 

Solution 2:

public boolean isScramble(String s1, String s2) {
    if(s1.equals(s2)) return true; // base case
    if(!isAnagram(s1, s2)) return false;
    int n = s1.length();
    for(int i = 1; i < n;i++){  
        String s1pre = s1.substring(0,i);  
        String s1suf = s1.substring(i);  
        String s2pre = s2.substring(0, i);  
        String s2suf = s2.substring(i);  
        boolean result = isScramble(s1pre, s2pre) && isScramble(s1suf, s2suf);  
        if(!result){  
            s2pre = s2.substring(0, s1.length() - i);  
            s2suf = s2.substring(s1.length() - i);  
            result = isScramble(s1pre, s2suf) && isScramble(s1suf, s2pre);  
        }  
        if(result) return true;  
    }  
    return false;  
}

private boolean isAnagram(String s1, String s2) {
    if(s1.length() != s2.length()) return false; 
    int[] table = new int[256];
    int n = s1.length();
    for(int i=0; i<n; i++) {
        table[s1.charAt(i)]++;
    }
    for(int i=0; i<n; i++) {
        char c = s2.charAt(i);
        if(--table[c] < 0) return false;
    }
    return true;
}