<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>

'''
TestCase:
1. abc -> False
2. abcabcabc ->True # so we cannot use divide it in half
3. a -> False
4. aabb -> False
# use sliding window, but it will change the window size
'''
class Solution:
    def repeatedSubstringPattern(self, s: str) -> bool:
        subString = ""
        
        for i,c in enumerate(s):
            
            if subString!="" and c == subString[0]:

                theSame =True
                
                for j in range(i, len(s), len(subString)):

                    if s[j:j+len(subString)]!=subString:
                        theSame = False
            
                if theSame:
                    return True
            subString+=c
        return False
                
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