<p>Given two strings <code>a</code> and <code>b</code>, return <em>the minimum number of times you should repeat string </em><code>a</code><em> so that string</em> <code>b</code> <em>is a substring of it</em>. If it is impossible for <code>b</code>​​​​​​ to be a substring of <code>a</code> after repeating it, return <code>-1</code>.</p>

<p><strong>Notice:</strong> string <code>&quot;abc&quot;</code> repeated 0 times is <code>&quot;&quot;</code>, repeated 1 time is <code>&quot;abc&quot;</code> and repeated 2 times is <code>&quot;abcabc&quot;</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> a = &quot;abcd&quot;, b = &quot;cdabcdab&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> We return 3 because by repeating a three times &quot;ab<strong>cdabcdab</strong>cd&quot;, b is a substring of it.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> a = &quot;a&quot;, b = &quot;aa&quot;
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= a.length, b.length &lt;= 10<sup>4</sup></code></li>
	<li><code>a</code> and <code>b</code> consist of lowercase English letters.</li>
</ul>

class Solution {
public:
    int repeatedStringMatch(string a, string b) {
        int size1 = a.size(), size2 = b.size();
        string temp = a;
        int addition = (size2 / size1) + 1;
        int repeated = addition + 1;
        for (int i = 0; i < addition; i++) {
            a += temp;
        }
        // cout << a << " " << repeated<<endl;
        int possible = false;
        for (int i = 0; i < temp.size(); i++) {
            string s1 = a.substr(i, size2);
            if (s1 == b) {
                // cout<<"i "<<i<<endl;
                int cnt = (a.size() - (i + size2));
                // cout<<cnt<<endl;
                repeated = repeated - (cnt / size1);
                possible = true;
                break;
            }
        }
        if (possible == false) {
            return -1;
        }
        return max(1, repeated);
    }
};
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