<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 odd(string s){
        string current = "";
        string ans ="";
        for(int i = 0 ;i <s.length() ; i++){
        int j= i;
        int k =i;
        while(  j >= 0 && k < s.length() && s[j] == s[k]){
            current = s.substr(j , k - j +1);
            j--;
            k++;
            if(current.length() > ans.length()){
                ans = current;
            }
        }
        }
        return ans;
    }
    string even(string s){
        string current = "";
        string ans ="";
        for(int i = 0 ;i <s.length() ; i++){
        int j= i;
        int k =i+1;
        while(  j >= 0 && k < s.length() && s[j] == s[k]){
            current = s.substr(j , k - j +1);
            j--;
            k++;
            if(current.length() > ans.length()){
                ans = current;
            }
        }
        }
        return ans;
    }
    string longestPalindrome(string s) {
     string ans1 = odd(s);
     string ans2 =even(s);
     string ans;
    if(ans1.length() > ans2.length()){
        ans = ans1;
    }else{
        ans = ans2;
    }
        return ans;    
    }
};
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