<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) {
        int n=s.length();
        // if(n%2) return false;
        
        vector<string>vec;
        string tmp;
        // all possible substring patterns, 1st letter vala char toh always include krna hoga, else bich mein se possible nhi 
        for(int i=0;i<n;i++){
            tmp+=s[i];
            if(tmp.length()>n/2) break;
            vec.push_back(tmp);
        }
        for(auto &sub:vec){
            // cout<<sub<<endl;
            int len=sub.length();
            int i=0;
            // bool found=false;
            while(i+len<=n){
                string str=s.substr(i, len);
                if(str!=sub) break;
                i+=len;
            } 
            if(i==n) return true;
        }
        return false;
    }
};
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