<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 check(string x, string s)
    {
        int n1 = x.length();
        int n2 = s.length();
        int i;

        for(i=0; i+n1-1<n2; i+=n1)
        {
            for(int j=0; j<n1; j++)
            {
                if(s[j+i] != x[j])
                {
                    return false;
                }
            }
        }

        if(i < n2)
        {
            return false;
        }

        return true;
    }

    string gcdOfStrings(string str1, string str2) {
        int n1 = str1.length();
        int n2 = str2.length();

        if(n2 > n1)
        {
            return gcdOfStrings(str2, str1);
        }

        for(int i=str2.length(); i>=1; i--)
        {
            if(check(str2.substr(0,i), str1) && check(str2.substr(0,i), str2))
            {
                return str2.substr(0,i);
            }
        }

        return "";
    }
};
31800