<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(std::string s) {
        
        int j {1};
        int counter {0};
        while (j < s.size() + 1){
            
            for(int i {0}; i < s.size(); i = i + j){
                // std::cout<< "s.substr(" << i << " , " << j << "): " << s.substr(i , j) << std::endl;
                // std::cout<< "s.substr(" << i + j << ", " << j << "): " << s.substr(i + j , j) << std::endl;
                if ( s.substr(i , j).size() > s.substr(i + j , j).size() ){
                    return false;
                }
                if (s.substr(i , j) == s.substr(i + j , j)){
                    counter++;
                    // std::cout<< "counter: " << counter << std::endl;
                    if (counter ==  std::ceil((double(s.size()) / double(j)) - 1)){
                        return true;
                    }
                }else{
                    j++;
                    counter = 0;
                    break;
                }
            }

        }
        return false;
    }
};
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