<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> computeLps(const string& str) {
        int n = str.length();
        int i = 1, len = 0;
        vector<int> lps(n, 0);

        while (i < n) {
            if (str[i] == str[len]) {
                len++;
                lps[i] = len;
                i++;
            } else {
                if (len != 0) {
                    len = lps[len - 1];
                } else {
                    i++;
                }
            }
        }
        return lps;
    }

    int repeatedStringMatch(string a, string b) {
        string str = "";
        if (a.length() <= 100) {
            for (int i = 0; i <= 10000; i++) {
                str += a;
            }
        } else if (a.length() > 100) {
            for (int i = 0; i <= 100; i++) {
                str += a;
            }
        }

        vector<int> lps = computeLps(b);

        int n = str.length();
        int m = b.length();
        int i = 0, j = 0;
        int ans = 1000000;

        while (i < n) {
            if (str[i] == b[j]) {
                i++;
                j++;
            }
            if (j == m) {
                ans = min(ans, i);
                break;
                // j = lps[j - 1];
            } else if (i < n && str[i] != b[j]) {
                if (j != 0) {
                    j = lps[j - 1];
                } else {
                    i++;
                }
            }
        }

        if (ans == 1000000) {
            return -1;
        }

        vector<int> arr;
        int cnt = a.length();
        int res = cnt;
        for (int k = 0; k < 1000000; k++) {
            arr.push_back(res);
            res += cnt;
        }

        for (int k = 0; k < arr.size(); k++) {
            if (arr[k] >= ans) {
                return k + 1;
            }
        }

        return -1;
    }
};
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