<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 ok(string a, string b){
        int m = a.size();
        int n = b.size();
        m/=n;
        string temp = "";
        while(m--){
            temp += b;
        }
        return a==temp;
    }
    string gcdOfStrings(string str1, string str2) {
        int n = str1.size();
        int m = str2.size();
        if(m<n){
            swap(str1, str2);
            swap(m,n);
        }
        string ans = "";
        for(int i=0;i<n;i++){
            if(n%(i+1)==0 && m%(i+1)==0 && ok(str2, str1.substr(0,i+1)) && ok(str1, str1.substr(0,i+1)))
                ans = str1.substr(0,i+1);
        }
        return ans;
    }
};
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