Given a string s, return the longest 
Example 1:
Input: s = "babad" Output: "bab" Explanation: "aba" is also a valid answer.
Example 2:
Input: s = "cbbd" Output: "bb"
Constraints:
- 1 <= s.length <= 1000
- sconsist of only digits and English letters.
Code(C++):-
class Solution {public:    string longestPalindrome(string s) {        int n=s.size();        int dp[n][n];        memset(dp,0,n*n*sizeof(int));        for(int i=0;i<n;i++)            dp[i][i]=1;        int maxlength=1;        int starting =0;        for(int i=0;i<n-1;i++)        {            if(s[i]==s[i+1])            {                starting=i;                dp[i][i+1]=1;                maxlength=2;            }        }                // for substring greater than 2        for(int k=3;k<=n;k++)        {            for(int i=0;i<=n-k;i++)            {                int j=i+k-1;                if(dp[i+1][j-1] and s[i]==s[j])                {                    if(k>maxlength)                    {                        maxlength=k;                        starting=i;                    }                    dp[i][j]=1;                                    }            }        }        string ans="";        for(int i=starting;i<starting+maxlength;i++)        {            ans+=(s[i]);        }        return ans;    }};
class Solution {
public:
    string longestPalindrome(string s) {
        int n=s.size();
        int dp[n][n];
        memset(dp,0,n*n*sizeof(int));
        for(int i=0;i<n;i++)
            dp[i][i]=1;
        int maxlength=1;
        int starting =0;
        for(int i=0;i<n-1;i++)
        {
            if(s[i]==s[i+1])
            {
                starting=i;
                dp[i][i+1]=1;
                maxlength=2;
            }
        }
        // for substring greater than 2
        for(int k=3;k<=n;k++)
        {
            for(int i=0;i<=n-k;i++)
            {
                int j=i+k-1;
                if(dp[i+1][j-1] and s[i]==s[j])
                {
                    if(k>maxlength)
                    {
                        maxlength=k;
                        starting=i;
                    }
                    dp[i][j]=1;
                }
            }
        }
        string ans="";
        for(int i=starting;i<starting+maxlength;i++)
        {
            ans+=(s[i]);
        }
        return ans;
    }
};
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