<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(const string& s1, const string& s2)
    {
        int ans = 0;
        int len = 1, n = s1.size(), m = s2.size();
        
        while(len <= n && len <= m){
            int i=0, j=0, k = 0;
            bool possible = true;

            if(n%len != 0 || m%len != 0){
                len++;
                continue;
            }

            while(i < n || j < m){
                if(i < n && s1[i] != s1[k]){
                    possible = false;
                    break;
                }
                else if(j < m && s2[j] != s1[k]){
                    possible = false;
                    break;
                }

                i++;
                j++;
                k++;

                if(i == n && k != len){
                    possible = false;
                    break;
                }
                if(j == m && k != len){
                    possible = false;
                    break;
                }

                if(k == len){
                    k = 0;
                }
            }

            //cout<<i<<" "<<j<<" "<<k<<" "<<len<<"\n";
            if(possible){
                ans = len;
            }

            len++;
        }

        return s1.substr(0, ans);
    }
};
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