<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) {
        // Step 1: Find the smaller of the two strings
        string minStr = (str1.length() < str2.length()) ? str1 : str2;

        // Step 2: Try each substring of the smaller string, starting from the longest
        for (int len = minStr.length(); len > 0; --len) {
            string candidate = minStr.substr(0, len);  // Possible divisor

            // Step 3: Check if this candidate can divide both strings
            if (canDivide(str1, candidate) && canDivide(str2, candidate)) {
                return candidate;  // This is the largest common divisor
            }
        }

        // Step 4: If no common divisor found, return empty string
        return "";
    }

private:
    // Helper function to check if string 's' can be formed by repeated concatenations of 't'
    bool canDivide(const string& s, const string& t) {
        int n = s.length();
        int m = t.length();
        
        // If length of s is not a multiple of length of t, t can't divide s
        if (n % m != 0) {
            return false;
        }

        // Check if repeating t forms the entire string s
        string repeatedT = "";
        for (int i = 0; i < n / m; ++i) {
            repeatedT += t;
        }

        return repeatedT == s;
    }
};
16635