<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>

#include <unordered_set>
#include <string>

class Solution {
public:
    int repeatedStringMatch(string a, string b) {
        // Check if every character in `b` exists in `a` at least once
        unordered_set<char> a_set(a.begin(), a.end());
        for (char c : b) {
            if (a_set.find(c) == a_set.end()) {
                return -1; // If any character in `b` is not in `a`, return -1
            }
        }

        // Now proceed with concatenating `a` to itself until `b` can be a substring
        string a1 = a;
        int count = 1;
        int len1 = a1.size();
        int len2 = b.size();

        // Continue to concatenate `a` until its length is at least the length of `b`
        while (len1 < len2) {
            a1 += a;
            count++;
            len1 = a1.size();
        }

        // Check if `b` is already a substring
        for (int i = 0; i <= a1.size() - b.size(); i++) {
            if (b == a1.substr(i, b.size())) {
                return count;
            }
        }

        // If not found, add one more repetition of `a` and check again
        a1 += a;
        count++;
        for (int i = 0; i <= a1.size() - b.size(); i++) {
            if (b == a1.substr(i, b.size())) {
                return count;
            }
        }

        // If `b` is still not a substring, return -1
        return -1;
    }
};

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