<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:

    // cheapest does divide check
    // does substring*len(string)/len(substring) = string

    // need a find longest repetitive string check
    // string and all of string[i:j] for j-i e {0:length}
    // once we find one, use it

    // find longest string for both
    // if they are the same, return
    // if they are different, compare substrings of each length until one does divide for both

    bool doesDivide(string& str, string subStr) {
        size_t lengthStr = str.length();
        size_t lengthSubStr = subStr.length();

        if (lengthStr%lengthSubStr != 0) {
            return false;
        }

        string potentialString = "";
        for (int i = 0; i < lengthStr/lengthSubStr; ++i) {
            potentialString += subStr;
        }
        return potentialString.compare(str)==0;
    }

    string gcdOfStrings(string str1, string str2) {
        string* shorter_string_pointer;
        string* longer_string_pointer;

        if (str1.length() < str2.length()) {
            shorter_string_pointer = &str1;
            longer_string_pointer = &str2;
        } else {
            shorter_string_pointer = &str2;
            longer_string_pointer = &str1;
        }

        for (int i=shorter_string_pointer->length(); i > 0; --i) {
            if (doesDivide(*longer_string_pointer, (shorter_string_pointer)->substr(0,i)) && doesDivide(*shorter_string_pointer, (shorter_string_pointer)->substr(0,i))) {
                return (shorter_string_pointer)->substr(0,i);
            }
        }

        return "";
    }
};
17056