<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 len=a.length();
        int lenb=b.length();
        for(int i=0;i<=len-lenb;i++){
            //cout<<a.substr(i,lenb)<<endl;
            if(a.substr(i,lenb)==b) return 1;
        }
        int ans=1e9;
        for(int i=0;i<=min(len,lenb);i++){//length of substring of b
            int cnt=0;
            if(a.substr(len-i,i)!=b.substr(0,i)) continue;//starting part of b not equal to end part of a
            if(i!=0) cnt++; //agr length of left substring is 0 toh count ++ nhi krna
            if(i==lenb){
                ans=min(ans,cnt);
                continue;
            }
            int j=i;//now the index
            while(1){
                if(lenb-j<len) break;
                string y=b.substr(j,len);
                if(y!=a) break;
                j=j+len;
                cnt++;
            }
            if(lenb-j>=len) continue;
            if(a.substr(0,lenb-j)!=b.substr(j,lenb-j)) continue;
            if(lenb!=j) cnt++;
            ans=min(ans,cnt);
        }
        if(ans==1e9) return -1;
        return ans;
    }
};
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