<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) {
        string curr_perfix = "", g_perfix = "";
        int i = 1;
        while (i <= str2.length()) {
            curr_perfix = str2.substr(0, i);
            int prefix_len = curr_perfix.length(), j = 0;
            if (str1.length() % prefix_len != 0 ||
                str2.length() % prefix_len != 0) {
                i++;
                continue;
            }
            bool matches = true;
            while (j < str1.length()) {
                if (str1.substr(j, prefix_len) != curr_perfix) {
                    matches = false;
                    break;
                }
                j += prefix_len;
            }
            if (matches) {
                j = 0;
                while (j < str2.length()) {
                    if (str2.substr(j, prefix_len) != curr_perfix) {
                        matches = false;
                        break;
                    }
                    j += prefix_len;
                }
            }
            if (matches) {
                g_perfix = curr_perfix;
            }

            i++;

        }
        return g_perfix;
    }
};
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