<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:
    bool isgcd(string &pre, string &tar){
        for(int i = 0 ; i < tar.size() ;i+=pre.size()){
            if(tar.substr(i,pre.size()) != pre)return false;
        }
        return true;
    }
  
    string gcdOfStrings(string str1, string str2) {
        if (str1[0] != str2[0]) return "";

        string res = "";

        // Determine the shorter string to iterate over
        string shorter = str1.size() < str2.size() ? str1 : str2;
        string longer = str1.size() < str2.size() ? str2 : str1;

        // Iterate over the shorter string to find the GCD
        for (int i = 0; i < shorter.size(); i++) {
            string pre = shorter.substr(0, i + 1);
            // Check if this prefix can form both strings
            if (isgcd(pre, shorter) && isgcd(pre, longer)) {
                res = pre; // Update result with the valid prefix
            }
        }
        
        return res; // Return the longest valid prefix found
    }
};
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