LeetCode – Refresh – Scramble String

Notes:

1. Make a array to check whether they have same number of chars (s1[i] - 'a') and (s2[i] - 'a')

2. When finally check, the i start from 1.

3. It is scramble as comparing s1(0, i) with s2(size - i, i) and s1(i) with s2(0, size-i). Kind of symmetry.

 

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