<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 ans, string str1) {
        int n = ans.size();
        int m = str1.size();
        if (m % n != 0)
            return false;
        for (int i = 0; i < m; i += n) {
             if (str1.substr(i, n) != ans)return  false ;
        }
        return true;
    }
    string gcdOfStrings(string str1, string str2) {
        int m = str1.size();
        int n = str2.size();
      
        if (n > m)
            return gcdOfStrings(str2,str1);
        int i = 0;
        string ans = "";
        while (i < n) {
            ans = str2.substr(0,n-i);
            if (check(ans, str1)&&check(ans,str2)) {
                return ans;
            } else {
                i++;
            }
        }
        return "";
    }
};
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