[leetcode]87. Scramble String

题目链接：https://leetcode.com/problems/scramble-string/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.

The basic idea is to divide s1(s2) into two substrings with length k and len-k and check if the two substrings s1[0..k-1] and s1[k, len-1] are the scrambles of s2[0..k-1] and s2[k,len-1] or s2[len-k, len-1] and s2[0..len-k-1] via recursion. The straigtforward recursion will be very slow due to many repeated recursive function calls. To speed up the recursion, we can use an unordered_map isScramblePair to save intermediate results. The key used here is s1+s2, but other keys are also possible (e.g. using indices)

(96ms)

class Solution {public:    bool isScramble(string s1, string s2) {        unordered_map<string,bool> isScramblePair;        return dp_helper(isScramblePair,s1,s2);    }private:    bool dp_helper(unordered_map<string,bool>& isScramblePair,string s1,string s2)    {        int len=s1.size();        bool res=false;        if(len==0)            return true;        else if(len==1)            return s1==s2;        else        {            // checked before, return intermediate result directly            if(isScramblePair.count(s1+s2))                return isScramblePair[s1+s2];            if(s1==s2)                res=true;            else            {                for(int i=1;i<len && !res;i++)                {                    //check if s1[0..k-1] and s1[k, len-1] are the scrambles of s2[0..k-1] and s2[k,len-1]                    res=res || (dp_helper(isScramblePair,s1.substr(0,i),s2.substr(0,i)) && dp_helper(isScramblePair,s1.substr(i,len-i),s2.substr(i,len-i)));                    //check if s1[0..k-1] and s1[k, len-1] are the scrambles of s2[len-k, len-1] and s2[0..len-k-1]                    res=res || (dp_helper(isScramblePair,s1.substr(0,i),s2.substr(len-i,i)) && dp_helper(isScramblePair,s1.substr(i,len-i),s2.substr(0,len-i)));                }            }            //save the intermediate results            return isScramblePair[s1+s2]=res;        }    }};

The recursive version has exponential complexity. To further improve the performance, we can use bottom-up DP, which is O(N^4) complexity. Here we build a table isS[len][i][j], which indicates whether s1[i..i+len-1] is a scramble of s2[j..j+len-1].

(22ms)

class Solution {public:    bool isScramble(string s1, string s2) {        int sSize=s1.size(),len;        if(sSize==0)            return true;        if(sSize==1)            return s1==s2;        bool isS[sSize+1][sSize][sSize]={false};        for(int i=0;i<sSize;i++)            for(int j=0;j<sSize;j++)                isS[1][i][j]=s1[i]==s2[j];        for(len=2;len<=sSize;len++)            for(int i=0;i<=sSize-len;i++)                for(int j=0;j<=sSize-len;j++)                {                    for(int k=1;k<len && !isS[len][i][j];k++)                    {                        isS[len][i][j]=isS[len][i][j] || (isS[k][i][j] && isS[len-k][i+k][j+k]);                        isS[len][i][j]=isS[len][i][j] || (isS[k][i+len-k][j] && isS[len-k][i][j+k]);                    }                }        return isS[sSize][0][0];    }};

(6ms)

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