<p>We define <code>str = [s, n]</code> as the string <code>str</code> which consists of the string <code>s</code> concatenated <code>n</code> times.</p>

<ul>
	<li>For example, <code>str == [&quot;abc&quot;, 3] ==&quot;abcabcabc&quot;</code>.</li>
</ul>

<p>We define that string <code>s1</code> can be obtained from string <code>s2</code> if we can remove some characters from <code>s2</code> such that it becomes <code>s1</code>.</p>

<ul>
	<li>For example, <code>s1 = &quot;abc&quot;</code> can be obtained from <code>s2 = &quot;ab<strong><u>dbe</u></strong>c&quot;</code> based on our definition by removing the bolded underlined characters.</li>
</ul>

<p>You are given two strings <code>s1</code> and <code>s2</code> and two integers <code>n1</code> and <code>n2</code>. You have the two strings <code>str1 = [s1, n1]</code> and <code>str2 = [s2, n2]</code>.</p>

<p>Return <em>the maximum integer </em><code>m</code><em> such that </em><code>str = [str2, m]</code><em> can be obtained from </em><code>str1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s1 = "acb", n1 = 4, s2 = "ab", n2 = 2
<strong>Output:</strong> 2
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s1 = "acb", n1 = 1, s2 = "acb", n2 = 1
<strong>Output:</strong> 1
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s1.length, s2.length &lt;= 100</code></li>
	<li><code>s1</code> and <code>s2</code> consist of lowercase English letters.</li>
	<li><code>1 &lt;= n1, n2 &lt;= 10<sup>6</sup></code></li>
</ul>

#include <string>
#include <unordered_map>
using namespace std;

class Solution {
public:
    int getMaxRepetitions(string s1, int n1, string s2, int n2) {
        if (n1 == 0) return 0;

        unordered_map<int, pair<int, int>> mp;
        int s1_count = 0, s2_count = 0, index = 0;

        while (s1_count < n1) {
            for (char c : s1) {
                if (c == s2[index]) {
                    index++;
                    if (index == s2.size()) {
                        index = 0;
                        s2_count++;
                    }
                }
            }
            s1_count++;

            if (mp.find(index) != mp.end()) {
                auto [prev_s1_count, prev_s2_count] = mp[index];
                int cycle_s1_count = s1_count - prev_s1_count;
                int cycle_s2_count = s2_count - prev_s2_count;

                int remaining_s1_count = n1 - s1_count;
                s1_count += remaining_s1_count / cycle_s1_count * cycle_s1_count;
                s2_count += remaining_s1_count / cycle_s1_count * cycle_s2_count;
            } else {
                mp[index] = {s1_count, s2_count};
            }
        }

        return s2_count / n2;
    }
};

8000