<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>

#pragma GCC optimize("O2,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")

static const bool Booster = [](){
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return true;
}();
class Solution {
public:
    string gcdOfStrings(string s1, string s2) {
        string res;
        string curr;
        for (int i = 0; i < s1.size() && i < s2.size(); ++i) {
            if (s1[i] != s2[i]) return res;
            curr += s1[i];
            if (curr != s1.substr(s1.size() - i - 1, i + 1)) continue;
            if (curr != s2.substr(s2.size() - i - 1, i + 1)) continue;
            if (check(curr, s1) && check(curr, s2)) res = curr;
        }
        return res;
    }

    bool check(const string& prefix, const string& s) {
        if (s.size() % prefix.size() != 0) return false;
        for (int i = 0; i < s.size(); i += prefix.size()) {
            if (prefix != s.substr(i, prefix.size())) return false;
        }
        return true;
    }
};
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