<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 helper(string a, string b ,int k){
        int ans =0;
        int j=0;
        int i=k;
        while(j<b.length() ){
            //cout<< "i,j "<<i<<" "<<j<<endl;
            if(b[j]== a[i]){
                j++;
                i++;
                if(j==b.length())
                {
                    
                    ans++;
                    return ans;
                }
                if(i == a.length()){
                    ans++;
                    i = 0;
                }
            }else{
                return INT_MAX;
            }
       
     }
     
     return INT_MAX;
    }
    int repeatedStringMatch(string a, string b) {
        int i=0,j=0;
        i = a.find(b[0]);
        if(i == -1)
        return -1;
    int ans =INT_MAX;
     for(int k=0;k<a.size();k++){
        if(b[0] == a[k]){
             //cout<<"k="<<k<<endl;
            ans = min(ans,helper(a,b,k));
            //cout<<"ans = "<<ans<<endl;
        
        }
     }
     return ans == INT_MAX? -1:ans;
    }
};
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