<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) {
        // Ensure str1 is always the larger string, swap if necessary
        if (str1.length() < str2.length()) {
            swap(str1, str2);
        }
        
        // Initialize the answer as an empty string
        string ans = "";
        
        // Iterate over all possible substrings of the smaller string (str2)
        for (int len = 1; len <= str2.length(); len++) {
            string candidate = str2.substr(0, len);
            
            // Check if candidate divides both strings
            if (dividesString(str1, candidate) && dividesString(str2, candidate)) {
                ans = candidate;
            }
        }
        
        return ans;
    }
    
private:
    // Helper function to check if str can be constructed by repeating pattern
    bool dividesString(const string& str, const string& pattern) {
        int strLen = str.length();
        int patternLen = pattern.length();
        
        if (strLen % patternLen != 0) {
            return false;
        }
        
        for (int i = 0; i < strLen; i += patternLen) {
            if (str.substr(i, patternLen) != pattern) {
                return false;
            }
        }
        
        return true;
    }
};

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