<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:
    bool isPalindrome(int l, int r, string s){
        while(l < r){
            if(s[l] != s[r]){
                return false;
            }
            l++;
            r--;
        }
        return true;
    }

    string longestPalindrome(string s) {
        int n = s.length();
        int maxLen = 0;
            int maxL = 0;
            int maxR = 0;
        for(int i = 0; i < n; i++){
            int l = i;
            int r = i;
            
            while((l >= 0) && (r < n) && (s[l] == s[r])){
                if(isPalindrome(l, r, s)){
                    int len = r - l + 1;
                    if(len > maxLen){
                        maxLen = len;
                        maxL = l;
                        maxR = r;
                    }
                }
                l--;
                r++;
            }
            l = i;
            r = i + 1;
            while((l >= 0) && (r < n) && s[l] == s[r]){
                if(isPalindrome(l, r, s)){
                    int len = r - l + 1;
                    if(len > maxLen){
                        maxLen = len;
                        maxL = l;
                        maxR = r;
                    }
                }
                l--;
                r++;
            }
        }
        return s.substr(maxL, maxR - maxL + 1);
    }
};
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