<p>Given a string <code>s</code>, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abab&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> It is the substring &quot;ab&quot; twice.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aba&quot;
<strong>Output:</strong> false
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcabcabcabc&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> It is the substring &quot;abc&quot; four times or the substring &quot;abcabc&quot; twice.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    bool repeatedSubstringPattern(string s) {
        // sliding window approach

        int w = 1;
        int max_size = s.size() / 2;
        while(w <= max_size)
        {
            // std::cout << w << " " << max_size << endl;
            for(int i = 0; i+w <= s.size();i= i+w)
            {
                // std::cout << w << " " << max_size << endl;
                if(s.substr(i, w) == s.substr(i+w, w) )
                {
                    if(i + 2*w == s.size())
                    {
                        return true;
                    }
                    std::cout << i << " " << w << endl;
                }
                else
                {
                    break;
                }
            }
            w++;
        }
        return false;

    }
};
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