<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 gcdOfStrings(string str1, string str2) {
        vector <string> vec1;
        vector <string> vec2;
        string ans = "";
        string temp = "";
        string str3 = "";

        int minsize = min(str1.length(), str2.length());

        if (minsize == str1.length()) {
            str3 += str1;
        } else {
            str3 += str2;
        }


        for (int i = 0; i < str3.size(); i++) {
            temp += str3[i];
            vec1.push_back(temp);
        }

        for (int j = 0; j < vec1.size(); j++) {
            if (str1.size() % (j + 1) == 0) {
                vec2.push_back(vec1[j]);
            }
        }

        for (int k = 0; k < vec2.size(); k++) {
            string h = "";
            string t = "";
            while (t.length() < str1.size()) {
                t += vec2[k];
            }
            while (h.length() < str2.size()) {
                h += vec2[k];
            }
            if (t == str1 && h == str2) {
                ans = vec2[k];
            }
        }

        return ans;
    }
};
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