<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:
    string joinStr(string s, int i) {
        string res;
        while(i--) {
            res = res + s;
        }
        return res;
    }

    bool validStr(string s1, string s2, int i) {
        int size1 = s1.size(), size2 = s2.size();
        if(size1 % i != 0 or size2 % i != 0) return false;

        int m = size1 / i;
        int n = size2 / i;
        string base = s1.substr(0, i);
        return joinStr(base, m) == s1 and joinStr(base, n) == s2;
    }

    string gcdOfStrings(string str1, string str2) {
        int size1 = str1.size();
        int size2 = str2.size();
        for(int i = min(size1, size2) ; i>0 ; i--) {
            if(validStr(str1, str2, i)) return str1.substr(0, i);
        }
        return "";
    }
};
38119