longest palindromic substring recursive solution

Question

I am aware of solutions that uses the bottom up dynamic programing approach to solve this problem in O(n^2). I am specifically looking for a top down dp approach. Is it possible to achieve longest palindromic substring using a recursive solution?

Here is what I have tried but it fails for certain cases, but I feel I am almost on the right track.

#include <iostream>
#include <string>

using namespace std;

string S;
int dp[55][55];

int solve(int x,int y,int val)
{

    if(x>y)return val;
    int &ret = dp[x][y];
    if(ret!=0){ret = val + ret;return ret;}
    //cout<<"x: "<<x<<" y: "<<y<<" val: "<<val<<endl;
    if(S[x] == S[y])
        ret = solve(x+1,y-1,val+2 - (x==y));
    else
        ret = max(solve(x+1,y,0),solve(x,y-1,0));
    return ret;
}


int main()
{
    cin >> S;
    memset(dp,0,sizeof(dp));
    int num = solve(0,S.size()-1,0);
    cout<<num<<endl;
}

Answer 1

For this case:

if(S[x] == S[y])
    ret = solve(x+1,y-1,val+2 - (x==y));

it should be:

if(S[x] == S[y])
    ret = max(solve(x + 1, y - 1, val + 2 - (x==y)), max(solve(x + 1, y, 0),solve(x, y - 1, 0)));

Because, in case you cannot create a substring from x to y, you need to cover the other two cases.

Another bug:

if(ret!=0){ret = val + ret;return ret;}

you should return ret + val and not modify ret in this case.

The main problem is you store the final val into dp[x][y], but this is not correct.

Example:

acabc , for x = 1 and y = 1, val = 3, so dp[1][1] = 3, but actually, it should be 1.

Fix:

int solve(int x,int y)
{  
    if(x>y)return 0;
    int &ret = dp[x][y];
    if(ret!=0){return ret;}

    if(S[x] == S[y]){
        ret = max(max(solve(x + 1, y),solve(x, y - 1)));
        int val = solve(x + 1, y - 1);
        if(val >= (y - 1) - (x + 1) + 1)
            ret = 2 - (x == y) + val;
    }else
        ret = max(solve(x+1,y),solve(x,y-1));
    return ret;
}