Leetcode Longest Palindromic Subsequence problem solution

class Solution:
    def longestPalindromeSubseq(self, s):

        if not s:
            return 0
        res = [[0 for i in range(len(s))] for i in range(len(s))]
        return self.helper(s, 0, len(s) - 1, res)

    def helper(self, inp, s, e, res):
        if s == e:
            return 1
        if s > e:
            return 0
        if res[s][e]:
            return res[s][e]
        if inp[s] == inp[e]:
            res[s][e] = 2 + self.helper(inp, s + 1, e - 1, res)
        else:
            res[s][e] = max(self.helper(inp, s + 1, e, res),
                         self.helper(inp, s, e - 1, res))
        return res[s][e]

class Solution {
    public int longestPalindromeSubseq(String s) {
       
        int n = s.length();
        StringBuilder b = new StringBuilder();         
        b.append(s);         
        b.reverse();
        return solve(s, b.toString(), n, n);
    
    }
    
  int solve(String str1, String str2, int n, int m){
        
        int[][] dp = new int[n + 1][m + 1];
        //initialize
        for(int i = 0; i < n + 1; i++){
            for(int j = 0; j < m + 1; j++){
                if(i == 0 || j == 0) dp[i][j] = 0;
            }
        }
        //main code
        for(int i = 1; i < n + 1; i++){
            for(int j = 1; j < m + 1; j++){
                if(str1.charAt(i - 1) == str2.charAt(j - 1)){
                    dp[i][j] = 1 + dp[i - 1][j - 1];
                }else{
                    dp[i][j] = Math.max(dp[i - 1][j],dp[i][j - 1]);
                }
            }
        }
        return dp[n][m];
    }
}

class Solution {
public:
    int longestPalindromeSubseq(string s) {
        int n = s.size();
        vector<vector<int>> dp(n+1,vector<int>(n+1,0));
        for(int i=1;i<dp.size();i++)
        {
            for(int j=1;j<dp.size();j++)
            {
                if(s[i-1] == s[n-j])
                {
                    dp[i][j] = 1 + dp[i-1][j-1];
                }else{
                    dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }
        return dp[n][n];
    }
};