<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:
    int gcd(int a, int b)
    {
        int e = min(a, b);
        for(int i = e; i > 0; i--)
        {
            if(a % i == 0 && b % i == 0) return i;
        }
        return 1;
    }
    string gcdOfStrings(string str1, string str2)
    {
        string out = "";
        int len = gcd(str1.length(), str2.length());
        for(int i = 0; i < len; i++)
        {
            if(len % (i+1) != 0) continue;

            string o = str1.substr(0, i + 1);
            string s1, s2;
            for(int j = 0; j < str1.length() / (i+1); j++)
            {
                s1 += o;
            }
            for(int j = 0; j < str2.length() / (i+1); j++)
            {
                s2 += o;
            }
            if(s1 != str1 || s2 != str2)
            {
                //break;
            }
            else
            {
                out = o;
            }
        }
        return out;
    }
};
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