<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>

#include <iostream>
#include <string> 

using namespace std;

class Solution {
public:

    bool isDivisor(const string& substr, const string& str){

        string combinedStr = ""; 
        int lengthDiv = str.length() % substr.length();

        if (lengthDiv != 0 ){
            return false;
        }

        for (int i=0; i < (str.length()/substr.length()); ++i){
            combinedStr += substr;
        }

        if (combinedStr == str) {
            return true;
        }
        
        return false;
    }

    string gcdOfStrings(string str1, string str2) {

        string candidate = "";
        string lesserWord;

        if (str1.length() >= str2.length()){
            lesserWord = str2;
        }
        else {
            lesserWord = str1;
        }

        for (int i = 0; i<lesserWord.size(); ++i){
            string substr = lesserWord.substr(0, i+1);

            if (isDivisor(substr, str1) && isDivisor(substr, str2)){

                candidate = substr;
            }
        }
        
        
        return candidate;



        
    }
};
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