<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 n = str1.length();
        int m = str2.length();
        if(n<m){
            swap(str1, str2);
            swap(n, m);
        }
        string ans = "";
        for(int i = 1;i<=m;i++){
            string temp = str1.substr(0, i);
            int j = 0;
            bool check = true;
            while(j<m){
                if(j+i>m){
                    check = false;
                    break;
                }
                if(temp!=str2.substr(j, i)){
                    check = false;
                    break;
                }
                j+=i;
            }
            if(!check){
                continue;
            }
            int k = 0;
            bool check2 = true;
            while(k<n){
                if(k+i>n){
                    cout<<"yaha hu"<<endl;
                    check2 = false;
                    break;
                }
                if(temp!=str1.substr(k, i)){
                    check2 = false;
                    break;
                }
                k+=i;
            }
            if(!check2){
                continue;
            }
            ans = temp;
        }
        return ans;
    }
};
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