<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:
    bool findMatch(string& text, string& match) {
        int n = match.length();
        vector<int> lsp(n, 0);
        int j = 0, i = 1;

        while(i < n) {
            if(match[i] == match[j]) {
                lsp[i++] = ++j; 
            } else if(j == 0) {
                i++;
            } else {
                j = lsp[j-1];
            }
        }

        i = 0, j = 0;
        while(i < text.length()) {
            if(text[i] == match[j]) {
                i++;
                j++;
            } else if(j == 0) {
                i++;
            } else {
                j = lsp[j-1];
            }
            if(j == n) return true;
        }

        return false;
    }

    int repeatedStringMatch(string a, string b) {
        string aCopy = a;
        int res = 1;
        while(a.length() < b.length()) {
            a += aCopy;
            res++;
        }

        int count = 0;

        do {
            if(findMatch(a, b)) return res;
            a += aCopy;
            res++;
            count++;
        } while(a.length() <= 2 * b.length() || count == 1);

        return -1;
    }
};
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