<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 longest=0,start=0;
        int explored=0;
        queue<pair<int,int>> q;
        q.push(make_pair(0,s.length()));
        while(!q.empty())
        {
            int left=q.front().first;
            int right=q.front().second;
            //cout<<left<<" "<<right<<endl;
            q.pop();
            if(explored==s.length())
                break;
            if(ispalindrome(s,left,right))
            {
                longest=right-left;
                start=left;
                break;
            }
            if(right==s.length())
            {
                explored++;
                left=0;
                right=s.length()-explored;
            }
            else
            {
                left++;
                right++;
            }
            q.push(make_pair(left,right));
        }
        return s.substr(start,longest);
    }
    bool ispalindrome(string & s,int left,int right)
    {
        for(int i=0;i<(right-left)/2;i++)
            if(s[i+left]!=s[right-i-1])
                return false;
        return true;
    }
};
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