<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 n=s.size();
          vector<vector<bool>> dp(n, vector<bool> (n,0));
          string ans=""; int cnt=0;
          for(int i=0;i<n;i++){dp[i][i]=1;}
          for(int i=n-2;i>=0;i--){
            string temp="";
            temp+=s[i];
            for(int j=i+1;j<n;j++){
                temp+=s[j];
                if(s[i]==s[j]){
                    if((i+1)<=(j-1)){
                          dp[i][j]=dp[i+1][j-1];
                    }else{
                          dp[i][j]=1;
                    }
                    if(dp[i][j] && (j-i+1)>cnt){
                         cnt=j-i+1;
                         ans=temp;
                    }
                }
            }
          }
          if(cnt==0){return ans+s[0];}
          return ans;
    }
};
144040