[LeetCode]Scramble String

class Solution {

//O(n^4) DP with some cut off can pass the large judge

//O(n^2) recursion with some cut off can also pass the large judge 

public:

    bool isScramble(string s1, string s2) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        if(s1.size() != s2.size()) return false;

        int n = s1.size();

        if(n == 0) return true;

        vector<vector<vector<bool> > > f(n, vector<vector<bool> >(n, vector<bool>(n+1, false)));

        //initialize

        for(int i = 0; i < n; ++i)

        {

            for(int j = 0; j < n; ++j)

            {

                if(s1[i] == s2[j]) f[i][j][1] = true;

            }

        }

        //dp

        for(int len = 2; len <= n; ++len)

        {

            for(int i = 0; i < n; ++i)

            {

                if(i+len-1 >= n) break;

                for(int j = 0; j < n; ++j)

                {

                    if(j+len-1 >= n) break;

                    for(int k = 1; k < len; ++k)

                    {

                        if(i+k < n && j+k < n) f[i][j][len] = f[i][j][len] || (f[i][j][k] && f[i+k][j+k][len-k]);

                        if(j+len-k < n && i+k < n) f[i][j][len] = f[i][j][len] || (f[i][j+len-k][k] && f[i+k][j][len-k]);

                        if(f[i][j][len] == true) break;

                    }

                }

            }

        }

        return f[0][0][n];

    }

};