<p>Given a string <code>s</code>, return <em>the longest</em> <span data-keyword="palindromic-string"><em>palindromic</em></span> <span data-keyword="substring-nonempty"><em>substring</em></span> in <code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;babad&quot;
<strong>Output:</strong> &quot;bab&quot;
<strong>Explanation:</strong> &quot;aba&quot; is also a valid answer.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;cbbd&quot;
<strong>Output:</strong> &quot;bb&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s</code> consist of only digits and English letters.</li>
</ul>

class Solution {
public:
    string longestPalindrome(string s) {
        int len=s.length();
        int max_len=1;
        int start=0;
        vector<vector<int>>dp(len,vector<int>(len,0));
        for(int i=0;i<len;i++)
          {
            dp[i][i]=1;
          }
          for(int i=0;i<len-1;i++)
            {
                if(s[i]==s[i+1])
                  {
                    dp[i][i+1]=1;
                    max_len=2;
                    start=i;
                  }
            }
            for(int k=3;k<=len;k++)
               {
                  for(int i=0;i<=len-k;i++)
                      {
                        int j=i+k-1;
                        if(s[i]==s[j] && dp[i+1][j-1]==1)
                            {
                                dp[i][j]=1;
                                
                                int prev=max_len;
                                max_len=max(max_len,k);
                                if(prev!=max_len)
                                  {
                                    start=i;
                                  }
                            }
                      }
               }
            string my_str="";
              for(int i=start;i<start+max_len;i++)
                 {
                    my_str=my_str+s[i];
                 } 
            return my_str;
          
    }
};
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