<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 p = 31;
    const int N = 1e5+7, m = 1e9+7;
    vector<long long> power;
    int repeatedStringMatch(string a, string b) {
        int A = a.size(), B = b.size();
        string a1 = "";
        int c = 0;
        while(a1.size()<b.size()){
            a1 += a; c++;
        }
        if (a1.find(b) != string::npos) {
            return c;
        }
        a1 += a;
        c++;
        if (a1.find(b) != string::npos) {
            return c;
        }
        power.resize(N, 1);
        power[0] = 1;
        for(int i=1;i<N;i++){
            power[i] = (power[i-1]*p)%m;
        }
        vector<long long> hash(a1.size()+1,0);
        for(int i=0;i<a1.size();i++){
            hash[i+1] = (hash[i] + (a1[i] - 'a' + 1)*power[i])%m;
        }
        long long h_s = 0;
        for(int i=0;i<B;i++){
            h_s = (h_s + (b[i] - 'a' + 1)*power[i])%m;
        }

        for(int i=0;i+B-1<a1.size();i++){
            long long c_h = (hash[i+B] - hash[i] + m)%m;
            if(c_h == (h_s)){
                return c;
            }
        }
        return -1;
    }
};
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