<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 isvalid(string s1, string s2, int k){
        int l1=s1.length(),l2=s2.length();
        if(l1%k!=0 && l2%k!=0){
            return false;
        }
        string base=s1.substr(0, k);
        return (s1==join(base, l1/k) ) && (s2==join(base, l2/k));
    }
    string join(string base, int l){
        string s="";
        for(int i=l; i>0; i--){
            s+=base;
        }
        return s;
    }
    string gcdOfStrings(string str1, string str2) {
        int n=str1.length(), m=str2.length();
        for(int i=min(n,m); i>0; i--){
            if(isvalid(str1, str2, i)){
                return str1.substr(0, i);
            }
        }
        return "";
    }
};
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