<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 gcd(string str, string div)
    {
        if(str.size()%div.size() != 0)
        return 0;
        int s = div.size();
        for(int i=0; i<str.size(); i+=s)
        {
            if(str.substr(i, s) != div)
            return 0;
        }
        return 1;
    }
    string gcdOfStrings(string str1, string str2) {
        int a = str1.size(), b = str2.size();
        string smaller = "";
        smaller = (a>b) ? str2 : str1;
        string ans = "";
        int i = smaller.size();
        while(i>0)
        {
            string div = smaller.substr(0, i--);
            if(gcd(str1, div) && gcd(str2, div))
            return div;
        }
        return ans;
    }
};
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