<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:
    // Helper function to check if 't' divides 's'
    bool divides(const string &s, const string &t) {
        int len_s = s.length();
        int len_t = t.length();
        
        if (len_s % len_t != 0) {
            return false;  // If t cannot fit evenly into s
        }
        
        // Repeatedly concatenate t and check if it forms s
        string repeated_t = "";
        for (int i = 0; i < len_s / len_t; ++i) {
            repeated_t += t;
        }
        
        return repeated_t == s;
    }
    
    string gcdOfStrings(string str1, string str2) {
        int len1 = str1.length();
        int len2 = str2.length();
        
        // The possible largest divisor string can only be as long as the smaller string
        int minLen = min(len1, len2);
        
        // Try all possible lengths from the smallest string length down to 1
        for (int len = minLen; len >= 1; --len) {
            // Extract the first 'len' characters from both strings
            string candidate = str1.substr(0, len);
            
            // Check if this candidate divides both strings
            if (divides(str1, candidate) && divides(str2, candidate)) {
                return candidate;  // Found the largest common divisor string
            }
        }
        
        return "";  // If no common divisor string is found
    }
};

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