<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) {
        //int i=0;
        int l1 = str1.size();
        int l2 = str2.size();
        string divisor = "";
        //int dL = 1; // divisor length

        if (str1[0] != str2[0]) {
            return divisor;
        }
        else {
            divisor = l1>l2 ? str1:str2;
        }

        int dL = divisor.size();
        int flag = 0;
        while (divisor != "") {
            flag = 0;
            for (int i=0; i<max(l1,l2); i+=dL) {
                if (i<l1) {
                    if (str1.substr(i,dL) != divisor) {
                        flag = 1;
                        break;
                    }
                }
                if (i<l2) {
                    if (str2.substr(i,dL) != divisor) {
                        flag = 1;
                        break;
                    }
                }
            }
            if (flag == 1) {
                divisor.pop_back();
                dL = divisor.size();
            }
            else {
                return divisor;
            }
        }
        return divisor;
    }
};
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