<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) {
        int n1 = str1.size();
        int n2 = str2.size();

        // Find common prefix of both strings.
        int len = 0;
        while (len < n1 && len < n2) {
            if (str1[len] == str2[len]) len++;
            else break;
        }

        // Determines whether prefix of length len divides the given string.
        auto valid = [&](int len, string s) -> bool {
            if (s.size() % len != 0) {
                return false;
            }
            for (int i = 0; i < s.size(); i++) {
                if (s[i] != str1[i % len]) {
                    return false;
                }
            }
            return true;
        };

        while (len) {
            if (valid(len, str1) && valid(len, str2)) {
                return str1.substr(0, len);
            } else {
                len -= 1;
            }
        }

        return "";
    }

};
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