<p>The <strong>DNA sequence</strong> is composed of a series of nucleotides abbreviated as <code>&#39;A&#39;</code>, <code>&#39;C&#39;</code>, <code>&#39;G&#39;</code>, and <code>&#39;T&#39;</code>.</p>

<ul>
	<li>For example, <code>&quot;ACGAATTCCG&quot;</code> is a <strong>DNA sequence</strong>.</li>
</ul>

<p>When studying <strong>DNA</strong>, it is useful to identify repeated sequences within the DNA.</p>

<p>Given a string <code>s</code> that represents a <strong>DNA sequence</strong>, return all the <strong><code>10</code>-letter-long</strong> sequences (substrings) that occur more than once in a DNA molecule. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"
<strong>Output:</strong> ["AAAAACCCCC","CCCCCAAAAA"]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "AAAAAAAAAAAAA"
<strong>Output:</strong> ["AAAAAAAAAA"]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is either <code>&#39;A&#39;</code>, <code>&#39;C&#39;</code>, <code>&#39;G&#39;</code>, or <code>&#39;T&#39;</code>.</li>
</ul>

class Solution {
public:
    vector<string> findRepeatedDnaSequences(string s) {
        if(s.length()<=10)
            return {};
       vector<string> subsets;
       vector<string> ans;
       for(int i=0;i<s.length()-9;i++)
       {
            string sub=s.substr(i,10);
            subsets.push_back(sub);
       }
       sort(subsets.begin(),subsets.end());
       int count=0;
       for(int i=0;i<subsets.size()-1;i++)
       {
           
            if((subsets[i]==subsets[i+1]))
            {
                count++;
            }
            else
            {
                if(count>0)
                {
                    ans.push_back(subsets[i]);
                    count=0;
                }
            }
            
        }
        if (count > 0) {
            ans.push_back(subsets.back());
        }
       
       return ans; 
    }
};
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