<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) {
        if (s.size() <= 1) {
            return false;
        }
        for (int len = 2; len < s.size();) {
            if (FoundForThisLength(s, len)) {
                return true;
            }

            len++;
            while (s.size() % len != 0) {
                len++;
            }
        }
        return FoundForThisLength(s, 1);
    }

private:
    bool FoundForThisLength(string s, size_t length) {
        string sub_str(s, 0, length);
        for (int i = 0; i < s.size(); i += length) {
            if (s.find(sub_str, i) != i) {
                return false;
            }
        }
        return true;
    }
};
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