[Leetcode] Scramble String

Scramble String

 

AC Rate: 318/1523

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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.

递归的来搞。比较粗暴。

class Solution {
public:
    bool isScramble(string s1, string s2) {
        int n = s1.size();
        if (!isSameChar(s1, s2)) return false;
        if (s1 == s2) return true;
        for (int i = 1; i < n; i++) {
            string s11 = s1.substr(0, i);
            string s12 = s1.substr(i);
            string s21 = s2.substr(0, i);
            string s22 = s2.substr(i);
            if (isScramble(s11, s21) && isScramble(s12, s22)) return true;
            s21 = s2.substr(0, n - i);
            s22 = s2.substr(n - i);
            if (isScramble(s11, s22) && isScramble(s12,s21)) return true;
        }
        return false;
    }
    
    bool isSameChar(string a, string b) {
        int c[256] = {0};
        for (int i = 0; i < a.size(); i++) {
            c[a[i]]++;
        }
        for (int i = 0; i < b.size(); i++) {
            if (--c[b[i]] < 0) return false;
        }
        return true;
    }
};

 

dp[i][j][k] means : s1[i~i+k], s2[j~j+k] is scramble.

        // dp[i][j][k] = 

        //          (dp[i][j][t] && dp[i+t][j+t][k-t]) || 

        //          (dp[i][j+k-t][t] &&dp[i+t][j][k-t]), (1<= t <= k-1)

Note:

loop need to begin from n - 1 to 0, if not, the result is not correct.

because s[0][0][n] need k from 1~n-1, so s[1][1][n-1] must be in the front of s[0][0][n].

 

class Solution {
public:
    bool isScramble(string s1, string s2) {
        int n = s1.size();
        bool dp[100][100][100] = {0};
        memset(dp, 0, sizeof(dp));
        // dp[i][j][k] 
        //  if true, s1[i~i+k], s2[j~j+k] match.
        //  else dismatch.
        // dp[i][j][k] = 
        //          (dp[i][j][t] && dp[i+t][j+t][k-t]) || 
        //          (dp[i][j+k-t][t] &&dp[i+t][j][k-t]), (1<= t <= k-1)
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                dp[i][j][0] = true;
                dp[i][j][1] = (s1[i] == s2[j]);
            }
        }

        for (int i = n - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                for (int k = 2; (i+k <= n) && (j+k <= n); k++) {
                    for (int t = 1; t < k; t++) {
                        if (dp[i][j][k]) break;
                        dp[i][j][k] |= (dp[i][j][t] && dp[i+t][j+t][k-t]) ||
                                        (dp[i][j+k-t][t] && dp[i+t][j][k-t]); 
                    }
                }
            }
        }
        return dp[0][0][n];;
    }
};