<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.length();

        // solution - 1 - brute force;

        // int i=0;
        // int maxLen = 0;
        // string pal;
        // for(i=0;i<n;i++){
        //     for(int j=i;j<n;j++){
        //         string currStr(s.begin()+i, s.begin()+j+1);
        //         string rev(currStr.rbegin(), currStr.rend());
        //         if(currStr == rev && (currStr.length() > maxLen)){
        //             pal = currStr;
        //             maxLen = currStr.length();
        //         }
        //     }
        // }
        // return pal;

        // solution - 2

        int i = 0;
        int maxLen = 0;
        string pal;
        for (i = 0; i < n; i++) {
            int j = n - 1;
            while (j >= i && ((j - i + 1) > maxLen)) {
                if (s[i] == s[j]) {
                    string currStr(s.begin() + i, s.begin() + j + 1);
                    string rev(currStr.rbegin(), currStr.rend());
                    if (currStr == rev) {
                        pal = currStr;
                        maxLen = currStr.length();
                    }
                }
                j--;
            }
        }
        return pal;
    }
};
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