Count The Repetitions
<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>
String; Dynamic Programming
function getMaxRepetitions(s1: string, n1: number, s2: string, n2: number): number {
    const n = s2.length;
    const d: number[][] = new Array(n).fill(0).map(() => new Array(2).fill(0));
    for (let i = 0; i < n; ++i) {
        let j = i;
        let cnt = 0;
        for (const c of s1) {
            if (c === s2[j]) {
                if (++j === n) {
                    j = 0;
                    ++cnt;
                }
            }
        }
        d[i] = [cnt, j];
    }
    let ans = 0;
    for (let j = 0; n1 > 0; --n1) {
        ans += d[j][0];
        j = d[j][1];
    }
    return Math.floor(ans / n2);
}
