Longest Palindromic substring

class Solution {
  public:
    string longestPalin (string s) {
        int n = s.size();
        vector <vector <bool>> dp(n,vector<bool>(n,false));

        int max_len = 1,start = 0;;
        for (int i = 0; i < n; ++i)   //one length substring will always be palindrome
            dp[i][i] = true;

        for (int i = 0; i < n - 1; ++i) {
            if (s[i] == s[i + 1]) {  //two length substring will be only palindrome if repeating eg aa,bb,cc
                dp[i][i + 1] = true;
                if(max_len<2){ start = i; max_len = 2; }
            }
        }

         /*so for abc:- a,b,c,ab,bc result are already in dp table,so for satisfying the palindromic
        condition inside elemetnts must be palindrome so if assume a==c then b must be palindrome for being abc
        palindrome that is already in dp table if that is palindrome or dp[i+1][j] where i and j is
        corner index of current substring is true then we will compare a and c,if same then abc is
        palindrome.
        eg2 sbcbs,find bcb is palindrome or not using dp table i.e dp[i+1][j-1] if yes,then compare
        ith and jt element,is same then it is palindrome else make dp[i][j]=false ie. not a palindrome
        */

        for (int k = 3; k <= n; ++k) {   //cheking palindromne length from 3 to n
            for (int i = 0; i < n - k + 1; ++i) {  //checking for every starting point
                int j = i + k - 1;    //ending point for current considered substring
                if (dp[i + 1][j - 1] && s[i] == s[j]) {
                    dp[i][j] = true;

                    if (k > max_len) {
                        start = i;
                        max_len = k;
                    }
                }
            }
        }
        return s.substr(start,max_len);
    }
};