<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 {
private:
    int BASE = 1000000;
public:
    int repeatedStringMatch(string A, string B) {
        if(A == B) return 1;
        int count = 1;
        string source = A;
        while(source.size() < B.size()){
            count++;
            source+=A;
        }
        if(source == B) return count;
        if(Rabin_Karp(source,B) != -1) return count;
        if(Rabin_Karp(source+A,B) != -1) return count+1;
        return -1;
    }
    int Rabin_Karp(string source, string target){
     //   if(source == "" or target == "") return -1;
        int m = target.size();
        
        int targetCode = 0;
        for(int i = 0;i<m;i++){
            targetCode += (target[i]);
        }
        int hashCode = 0;
        for(int i = 0;i<source.size();i++){
            hashCode += source[i];
            if(i<m-1) continue;
            if(i>=m){
                hashCode = (hashCode-source[i-m]);
            }
           
            if(hashCode == targetCode){
                if(source.substr(i-m+1,m) == target)
                    return i-m+1;
            }
        }
        return -1;
    }
};
19278