<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 {
private:
    bool check(int i, string& str1, string& str2) {
        string f1 = "";
        string f2 = "";
        string modStr1 = str1.substr(0, i);
        string modStr2 = str2.substr(0, i);
        int times1 = str1.length() / i;
        int times2 = str2.length() / i;
        for (int c = 0; c < times1; c++) {
            f1.append(modStr1);
        }
        for (int c = 0; c < times2; c++) {
            f2.append(modStr2);
        }
        return modStr1 == modStr2 && f1 == str1 && f2 == str2;
    }
public:
    string gcdOfStrings(string str1, string str2) {
        int len = 0;
        for (int i = 1; i <= min(str1.length(), str2.length()); i++) {
            if ((str1.length() % i) == 0 && (str2.length() % i) == 0) {
                if (check(i, str1, str2)) {
                    len = i;
                }
            }
        }
        return str1.substr(0, len);
    }
};
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