<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:
    bool divisable(string& s, string& t) {
        if (s.size() % t.size() != 0) return false;
        for (int i = 0; i < s.size(); i++) { 
            if (s[i] != t[i % t.size()]) return false;
        }
        return true;
    }

    string gcdOfStrings(string str1, string str2) {
        int minLen = min(str1.size(), str2.size());
        string res = "";
        for (int i = 0; i < minLen; i++) {
            auto sstr1 = str1.substr(0, i + 1);
            auto sstr2 = str2.substr(0, i + 1); 
            if (sstr1 == sstr2 && divisable(str1, sstr1) && divisable(str2, sstr2)) {
                res = sstr1;
            }
        } 
        return res;
    }
};
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