Palindrome Partitioning II

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.

class Solution {

public:

    int minCut(string s) 

    {     

        int n=s.length();        

        

        int* pali1=new int[n+1];

        int* pali2=new int[n+1];

        //1  3 5

        for(int i=0;i<n;i++) pali1[i]=1;

        for(int len=3;len<=n;len=len+2)

            for(int l=len/2;l<=n-1-len/2;l++)

                if(pali1[l]==len-2 && s[l-len/2]==s[l+len/2])

                    pali1[l]=len;

        //0 2 4

        for(int i=0;i<n;i++) pali2[i]=0;

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

            for(int l=len/2-1;l<=n-1-len/2;l++)

                if(pali2[l]==len-2 && s[l+1-len/2]==s[l+len/2])

                    pali2[l]=len;

        int** cutcache=new int*[n];

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

        {

            cutcache[i]=new int[n];

            for(int j=0;j<n;j++) cutcache[i][j]=-1;

        }

        return mincut(cutcache,s,0,n-1,pali1,pali2);

    }

    int mincut(int** cutcache,const string& s,int l,int r,const int* pali1,const int* pali2)

    {

        if(cutcache[l][r]!=-1) return cutcache[l][r];

        if(l>=r) return 0;

        

        if(check(s,l,r,pali1,pali2)) 

        {

            cutcache[l][r]=0;

            return 0;

        }

        int cut=r-l+1;

        for(int i=l;i<r;i++)

        if(check(s,l,i,pali1,pali2))

        {

            int newcut=1+mincut(cutcache,s,i+1,r,pali1,pali2);

            if(newcut<cut) cut=newcut;

        }

        cutcache[l][r]=cut;

        

        return cut;

    }

    bool check(const string& s,int l,int r,const int* pali1,const int* pali2)

    {

        int len=r-l+1;

        int mid=(l+r)/2;

        if(len%2==1 && pali1[mid]>=len) return true;

        if(len%2==0 && pali2[mid]>=len) return true;

        return false;

    }

};