<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:
    vector<int> getNextArr(string a) {
        vector<int> arr(a.size(), 0);
        if (a.size() > 1) 
            arr[0] = -1;
        if (a.size() > 2)
            arr[1] = 0;

        int cur = 0;
        for (int i = 1; i < a.size() - 1;) {
            if (a[i] == a[cur]) {
                arr[i + 1] = cur + 1;
                i++;
                cur++;
            } else if (cur > 0){
                cur = arr[cur];
            } else {
                i++;
            }
        }

        return arr;
    }

    bool isSubstr(string a, string b) {
        auto next = getNextArr(b);
        int i = 0;
        int cur = 0;
        while (i < a.size()) {
            if (a[i] == b[cur]) {
                cur ++;
                i ++;
            } else if (cur > 0) {
                cur = next[cur];
            } else {
                i++;
            }
            if (cur == b.size())
                return true;
        }
        return cur == b.size();
    }

    int repeatedStringMatch(string a, string b) {
        string final_a = "";
        int result = 0;
        
        while (final_a.size() < b.size()) {
            final_a += a;
            result += 1;
        }

        while (final_a.size() <= 10000) {
            // cout << final_a << endl;
            if (isSubstr(final_a, b))
                return result;
            final_a += a;
            result += 1;
        }
        
        return -1;
    }
};
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