<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) {
        string base, str, base_bkp;
        int str1_len = str1.length();
        int str2_len = str2.length();
        int base_len, b_len;
        bool match_found = false;
        if (str1.length() > str2.length()) {
            str = str1;
            base = str2;
            base_bkp = str2;
            base_len = str2_len;
        } else {
            base = str1;
            str = str2;
            base_bkp = str1;
            base_len = str1_len;
        }
        b_len = base_len;
        while (base_len > 0) {
            if (str.length()%base_len == 0) {
                int k = str.length()/base_len;
                int i;
                string base_k = "";
                for (i=0;i<k;i++) {
                    base_k += base;
                }
                if (str == base_k) {
                    if (b_len%base_len==0) {
                        k = b_len/base_len;
                        base_k = "";
                        for (i=0;i<k;i++) {
                            base_k += base;
                        }
                        if (base_k == base_bkp) {
                            match_found = true;
                            break;
                        }
                    }
                }
            }
            base = base.substr(0, base_len-1);
            base_len--;
        }
        if (match_found)
            return base;
        else
            return "";
    }
};
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