<p>For two strings <code>s</code> and <code>t</code>, we say &quot;<code>t</code> divides <code>s</code>&quot; if and only if <code>s = t + t + t + ... + t + t</code> (i.e., <code>t</code> is concatenated with itself one or more times).</p>

<p>Given two strings <code>str1</code> and <code>str2</code>, return <em>the largest string </em><code>x</code><em> such that </em><code>x</code><em> divides both </em><code>str1</code><em> and </em><code>str2</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> str1 = &quot;ABCABC&quot;, str2 = &quot;ABC&quot;
<strong>Output:</strong> &quot;ABC&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> str1 = &quot;ABABAB&quot;, str2 = &quot;ABAB&quot;
<strong>Output:</strong> &quot;AB&quot;
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> str1 = &quot;LEET&quot;, str2 = &quot;CODE&quot;
<strong>Output:</strong> &quot;&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= str1.length, str2.length &lt;= 1000</code></li>
	<li><code>str1</code> and <code>str2</code> consist of English uppercase letters.</li>
</ul>

class Solution {
public:
    set<string> solve(string & str){
        string temp="";
        set<string> ans;
        for(int i=0;i<str.length();i++){
            temp.push_back(str[i]);
            int flag=0;
            for(int j=i+1;j<str.length();j+=(i+1)){
                if(str.substr(j,i+1)!=temp){
                    flag=1;
                    break;
                }
            }
            if(flag==0){
                if(temp.size()>ans.size()){
                    ans.insert(temp);
                }
            }
        }
        
        return ans;
    }
    string gcdOfStrings(string str1, string str2) {
        
        set<string> t1=solve(str1);
        set<string> t2=solve(str2);
        string ans="";
        for(auto x:t1){
            if(t2.find(x)!=t2.end()){
                if(x.length()>ans.length()){
                    ans=x;
                }
            }
        }
        return ans;
//         string temp="";
//         if(t1.length()==t2.length()){
//             if(t1==t2){
//                 return t1;
//             }
//             return "";
//         }
        
    
//         if(t1.length()<t2.length()){
//             temp=t1;
//         }
//         else{
//             temp=t2;
//         }
//         for(int i=0;i<str.length();i++){
//             int flag=0;
//             for(int j=i+1;j<str.length();j+=(i+1)){
//                 if(str.substr(j,i+1)!=temp){
//                     flag=1;
//                     break;
//                 }
//             }
//             if(flag==0){
//                 if(temp.size()>ans.size()){
//                     ans=temp;
//                 }
//             }
//         }
        
    }
};
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