Permutation in String
<p>Given two strings <code>s1</code> and <code>s2</code>, return <code>true</code> if <code>s2</code> contains a <span data-keyword="permutation-string">permutation</span> of <code>s1</code>, or <code>false</code> otherwise.</p>

<p>In other words, return <code>true</code> if one of <code>s1</code>&#39;s permutations is the substring of <code>s2</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s1 = &quot;ab&quot;, s2 = &quot;eidbaooo&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> s2 contains one permutation of s1 (&quot;ba&quot;).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s1 = &quot;ab&quot;, s2 = &quot;eidboaoo&quot;
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s1.length, s2.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s1</code> and <code>s2</code> consist of lowercase English letters.</li>
</ul>
Hash Table; Two Pointers; String; Sliding Window
function checkInclusion(s1: string, s2: string): boolean {
    let need = 0;
    const cnt: number[] = Array(26).fill(0);
    const a = 'a'.charCodeAt(0);
    for (const c of s1) {
        if (++cnt[c.charCodeAt(0) - a] === 1) {
            need++;
        }
    }

    const [m, n] = [s1.length, s2.length];
    for (let i = 0; i < n; i++) {
        let c = s2.charCodeAt(i) - a;
        if (--cnt[c] === 0) {
            need--;
        }
        if (i >= m) {
            c = s2.charCodeAt(i - m) - a;
            if (++cnt[c] === 1) {
                need++;
            }
        }
        if (need === 0) {
            return true;
        }
    }
    return false;
}
