<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:
    string gcdOfStrings(string str1, string str2) {
        int n = str2.size();
        int n1 =str1.size();
        int max_no = 0;
        
        string result;
    
        
        for(int l=1 ;l<=n && l<=n1 ;l++ ){
            string pattern = str1.substr(0 ,l);
            string newstr ="";
            if(n1%l == 0 && n%l==0){
                int j=0 ;
                while(j < str1.size()/l){
                    newstr += pattern;
                    j++;
                }
                j=0;
            
            string newst ="";
            
            
            while(j < str2.size()/l){
                    newst += pattern;
                    j++;
                }

                if(str1 == newstr  && str2 == newst){
                if(max_no < pattern.size()){
                    result = pattern;
                    max_no =pattern.size();
                }
                }
            }
                   }


        
        return result;
           }
    
                   
        
    
};
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