<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>

#include <unordered_map>
class Solution {
public:
    int repeatedStringMatch(string a, string b) {
        // make a frequency map, storing all the characters in a, then check for b, if it doesn't contain some characters, return -1
        unordered_map <char,int> map;
        string a1 = a;
        for(int i=0;i<a1.size();i++){
            map[a1[i]]++;
        }
        for(int i=0;i<b.size();i++){
            if(map[b[i]] == 0) return -1;
        }
        // keep concatenating a to a, and check when b becomes a substring
        int count = 1;
        int len1 = a1.size();
        int len2 = b.size();
        while(len1 < len2){
            a1 += a;
            count++;
            len1 = a1.size();
        }
        // check if we have already gotten the 
        for(int i=0;i<=a1.size()-b.size();i++){
            if(b == a1.substr(i,b.size())){
                return count;
            }
        }
        // if we haven't gotten the substring
        a1 += a;
        count++;
        for(int i=0;i<=(a1.size()-b.size());i++){
            if(b == a1.substr(i,b.size())){
                return count;
                }
            }
        return -1;
    }
};
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