<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:
    string repeat(string str, int count){
        string result = "";
        for (int i = 0; i<count; i++){
            result += str;
        }
        return result;
    }
    bool repeatedSubstringPattern(string s) {
        unordered_map<string, int> store = {};
        string substring;
        int num_repeats;

        for (int j = 0; j<=s.size() / 2; j++){
            // store[s.substr(0, i)]++;
            substring = s.substr(0, j+1);
            if (s.size() % substring.size() == 0){
                num_repeats = s.size() / substring.size();
                if (num_repeats != 1 && repeat(substring, num_repeats) == s){ 
                    return true;
                }
            }
        }
        return false;

    }
};
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