<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) {
        string sub = "";
        sub += s[0];
        // 0 1 2 3
        // 4 - 1 = 3
        // 4 -2 = 2
        for (int i = 1; i <= s.size() - sub.size(); i++) {
            // cout << "i: " << i << endl;
            // cout << "sub: " << sub << endl;
            // cout << "s.substr: " << s.substr(i, sub.size()) << endl;
            if (sub == s.substr(i, sub.size())) {
                if (i == s.size() - sub.size())
                    return true;

                i += sub.size() - 1;
            } else {
                sub = s.substr(0, sub.size() + 1);
                i = sub.size() - 1;
            }
        }
        return false;
    }
};
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