longest palindromic substring recursive solution

Longest Palindrome using Recursion in Javascript:

const longestPalindrome = str => {
  if (str.length > 1){
    let [palindrome1, palindrome2] = [str, str];
    for (let i=0;i<Math.floor(str.length/2);i++) {
      if(str[i]!==str[str.length-i-1]) {
        palindrome1 = longestPalindrome(str.slice(0, str.length-1));
        palindrome2 = longestPalindrome(str.slice(1, str.length));
        break;
      }
    }
    return palindrome2.length > palindrome1.length ? palindrome2 : palindrome1;
  } else {
    return str;
  }
}

console.log(longestPalindrome("babababababababababababa"));

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;
}