<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) {
        string front = "",
               back = "",
               pattern = "";
        int total = 0,
            n = s.size(),
            curr = 0;
        unordered_map<string,int> mp;
        for(int i=0; i<n; i++){
            if(i>= n-1-i) break;
            front += s[i];
            back = s[n-1-i]+ back;
            if(front== back){
                total = i+1;
                curr = total;
                pattern = front;
                //break;
            }
        }
        if(pattern.size()< 1) return false;
        front = "";
        for(int j=pattern.size(); j<n; j++){
            front += s[j];
            if(front.size()== pattern.size()){
                if(front== pattern) front = "";
                else return false;
            }
        }
        return true;
    }
};
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