<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:
    int check(string s)
    {
        int l=s.length();
        for(int i=0;i<l/2;i++)
        if(s[i]!=s[l-i-1])
        return 0;
        return 1;
    }
    string longestPalindrome(string s) {
        string t=s;
        reverse(s.begin(),s.end());
        int m=s.length(),n=t.length(),maxLen=0,endIndex=0;;
        vector<vector<int>> dp(m+1,vector<int>(n+1,0));
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(s[i-1]==t[j-1])
                {
                    dp[i][j]=1+dp[i-1][j-1];
                    if(dp[i][j]>maxLen && check(s.substr(i - dp[i][j], dp[i][j]))) 
                    {
                        maxLen = dp[i][j];
                        endIndex = i - 1;
                    }
                }
                else
                dp[i][j]=0;
            }
        }
        return s.substr(endIndex - maxLen + 1, maxLen);
    }
};
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