<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 n=a.length();
        int m=b.length();
        int cnt=1;
        int pos=-1;
        string x=a;
        unordered_map<char,int>hash1;
        unordered_map<char,int>hash2;
        for(int i=0;i<n;i++){
            hash1[a[i]]++;
        }
        for(int i=0;i<m;i++){
            hash2[b[i]]++;
        }
        if(hash1.size()<hash2.size())return -1;



        while(n<=2*m){
            if(n<m){
                a=a+x;
                cnt++;
                n=a.length();
                continue;
            }
            pos=a.find(b);
             if(pos>=0)return cnt;
            else{
            a=a+x;
            cnt++;
         }
         n=a.length();
        
        }
        if(n>m){
            pos=a.find(b);
            if(pos>=0){
                return cnt;
            }
            a=a+x;
            cnt++;
            pos=a.find(b);
            if(pos>=0){
                return cnt;
            }

        }
    return -1;
    }
};
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