87. Scramble String

problem 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":

    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.

Example 1:

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

Example 2:

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

algorithm thought

判断是否Scramble，可以用递归的方式进行。如果一个字符串的左边和另一个字符串的左边是scramble，右边和右边是scramble。或者是，左右，右左是scramble字符串的话。那么这两个字符串就是scramble字符串。可以通过剪枝加快判断，首先判断两个字符串是否包含同样的字符，两个字符串中包含的字符不一样，就可以直接返回。虽然这里用了O(n)的时间，但是能剪去很多分支，是值得的。

code

class Solution {
public:
    bool isScramble(string s1, string s2) {
      //  cout<<s1<<':'<<s2<<' ';
        if(s1==s2)
            return true;
        int count[26]={0};
        int n=s1.size();
        for(int i=0;i<n;++i){
            count[int(s1[i]-'a')]++;
            count[int(s2[i]-'a')]--;
        }
        for(int i=0;i<26;++i)
            if(count[i]!=0)
                return false;
        for(int i=1;i<n;++i){
            if(isScramble(s1.substr(0,i),s2.substr(0,i))&&isScramble(s1.substr(i),s2.substr(i)))
                return true;
            if(isScramble(s1.substr(n-i),s2.substr(0,i))&&isScramble(s1.substr(0,n-i),s2.substr(i,n-i)))
                return true;
        }
        return false;
    }
};

algorithm analysis

每个函数正常情况下都能分出4个分支，每个分支中时间复杂度是线性的。最后时间复杂度不好定义。但是整个运行时间复杂度还是挺高的