Longest palindromic substring top down recursive approach

Question

I am trying to solve Longest palindromic substring on Leetcode. I a aware of solutions for this problem like expand around center or dynamic programming bottom up approach. For purely educational purposes I wanted to solve this in top down recursive manner. I am trying to find solution similar to what is described here or here. (problem is slighly different). I have this function:

private (int Start, int End) Longest(string s, int i, int j)

which takes string + start and end position of search. Tuple which is return is start and end of longest palindrom. I am trying to split into these cases:

1. if s[i] == s[j] investigate Longest(s, i+1, j-1)
2. Investigate Longest(s, i+1, j)
3. Investigate Longest(s, i, j - 1)
4. Return longest (max difference between returned start and end) from these three

Of course I am using Dictionary with tuple (int, int) as key (values of i and j) to remember all computed results to not compute them again.

Full code is below but it is very messy after few several iterations when I was trying to fix algorithm. I believe concreate code is not very important.

Code seems to be returning correct results but fail on Time Limit Exceeded on Leetcode. Is there correct fast recursive solution? I believe there should be as DP bottom up solution exists.

code:

private readonly IDictionary<(int, int), (int, int)> _mem = new Dictionary<(int, int), (int, int)>();

private (int Start, int End) Longest(string s, int i, int j) {
    if (i >= j) {
        return (i, j);
    }

    if (_mem.TryGetValue((i, j), out var ret)) {
        return ret;
    }

    var newI = i + 1;
    var newJ = j - 1;

    ValueTuple<int, int> removingTwo;

    if (s[i] == s[j])
    {
        removingTwo = Longest(s, newI, newJ);

        if (removingTwo.Item1 == newI && removingTwo.Item2 == newJ) {
            removingTwo.Item1--;
            removingTwo.Item2++;
        }
    }
    else {
        removingTwo = (1, 0);
    }

    var removingFirst = Longest(s, newI, j);
    var removingLast = Longest(s, i, newJ);  

    var mT = removingTwo.Item2 - removingTwo.Item1;
    var mF = removingFirst.End - removingFirst.Start;
    var mL = removingLast.End - removingLast.Start;

    var max = Math.Max(mT, mF);
    max = Math.Max(max, mL);

    ValueTuple<int, int> retVal;

    if (max == mT) retVal = removingTwo;
    else if (max == mF) retVal = removingFirst;
    else retVal = removingLast;

    _mem.Add((i, j), retVal);

    return retVal;

}

Edit: working bottom-up solution (copied from geegsforgeegs):

public string LongestPalindrome(string s) {
    if (s.Length == 0) return "";
    var table = new bool[s.Length, s.Length];
    var len = s.Length;
    for (int i = 0; i < len; i++) {
        table[i,i] = true;
    }

    var start = 0;
    var max = 1;
    for (int i = 0; i < len - 1; i++) {
        if (s[i] == s[i + 1]) {
            start = i;
            max = 2;
            table[i, i+1] = true;
        }
    }

    for (int k = 3; k <= len; ++k) { 

              // Fix the starting index 
        for (int i = 0; i < len - k + 1; ++i)  
        { 
            // Get the ending index of substring from 
            // starting index i and length k 
            int j = i + k - 1; 

            // checking for sub-string from ith index to 
            // jth index iff str.charAt(i+1) to  
            // str.charAt(j-1) is a palindrome 
            if (table[i + 1, j - 1] && s[i] == s[j]) { 
                table[i,j] = true; 

                if (k > max) { 
                    start = i; 
                    max = k; 
                } 
            } 
        } 
    } 

    return s.Substring(start, max);
}

Answer 1

Here's a recursive method in Python that passes the LeetCode test. It may be that they are looking for a constant space solution.

f(i, k) returns (l, j), the largest tuple of length l and its starting index, j. max in this instance is looking at the first element of the returned tuple, which is l, the palindrome's length.

def longestPalindrome(self, s):
  def f(i, k):
    return max(
      # next iteration
      f(i - 1, 1) if k < 2 and i > 0 else (-1,-1),
      f(i - 1, 2) if k < 2 and i > 0 and s[i-1] == s[i] else (-1, -1),

      # a larger palindrome than this one
      f(i - 1, k + 2) if i > 0 and i + k < len(s) and s[i-1] == s[i + k] else (-1, -1),

      # this one
      (k, i)
    )

  (l, j) = f(len(s) - 1, 1)
  return s[j:j+l]