<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>

class Solution:
    def getMaxRepetitions(self, s1: str, n1: int, s2: str, n2: int) -> int:
        recall = {}  # Dictionary to detect cycles
        s1_count, s2_count, index = 0, 0, 0

        while s1_count < n1:
            for char in s1:
                if char == s2[index]:
                    index += 1
                    if index == len(s2):
                        index = 0
                        s2_count += 1

            s1_count += 1

            # Detect cycle
            if index in recall:
                prev_s1_count, prev_s2_count = recall[index]
                cycle_s1_count = s1_count - prev_s1_count
                cycle_s2_count = s2_count - prev_s2_count

                # How many full cycles fit in the remaining `s1`
                remaining_s1 = (n1 - s1_count) // cycle_s1_count

                # Fast-forward the repetitions
                s1_count += remaining_s1 * cycle_s1_count
                s2_count += remaining_s1 * cycle_s2_count

            # Store the current state
            recall[index] = (s1_count, s2_count)

        return s2_count // n2

16500