<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 joinWords(int f, string s) {
        string ans = "";
        for (int i = 0; i < f; i++) {
            ans += s;
        }
        return ans;
    }
    bool helper(string str1, string str2, int l) {
        int len1 = str1.size();
        int len2 = str2.size();
        if (len1%l > 0 || len2%l > 0) {
            return false;
        }
        else {
            int mul_factor1 = len1/l; // 6 / 2 = 3 * (AB)
            int mul_factor2 = len2/l; // 4 / 2 = 2 * (AB)
            if (str1 == joinWords(mul_factor1, str1.substr(0, l)) && str2 == joinWords(mul_factor2, str1.substr(0, l))) {
                return true;
            }
            else {
                return false;
            }
        }
        
    }
    string gcdOfStrings(string str1, string str2) {
        // ABABAB 6%4 != 0
        // ABAB  4%4 = 0
        //have to be greeedy and start from last...(as we need largest)
        // so check ABAB, but str1 can not formed as it has to be divided
        // ABA , this str2 can not form 
        // AB....this can and this is answer
        int len1 = str1.size();
        int len2 = str2.size();
        for (int i = min(len1, len2); i >= 1; i--) { // loops runs from ABAB to AB
            if (helper(str1, str2, i)) {
                return str1.substr(0, i);
            }
        }
        return "";
    }
};
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