Find All Anagrams in a String
<p>Given two strings <code>s</code> and <code>p</code>, return an array of all the start indices of <code>p</code>&#39;s <span data-keyword="anagram">anagrams</span> in <code>s</code>. You may return the answer in <strong>any order</strong>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;cbaebabacd&quot;, p = &quot;abc&quot;
<strong>Output:</strong> [0,6]
<strong>Explanation:</strong>
The substring with start index = 0 is &quot;cba&quot;, which is an anagram of &quot;abc&quot;.
The substring with start index = 6 is &quot;bac&quot;, which is an anagram of &quot;abc&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abab&quot;, p = &quot;ab&quot;
<strong>Output:</strong> [0,1,2]
<strong>Explanation:</strong>
The substring with start index = 0 is &quot;ab&quot;, which is an anagram of &quot;ab&quot;.
The substring with start index = 1 is &quot;ba&quot;, which is an anagram of &quot;ab&quot;.
The substring with start index = 2 is &quot;ab&quot;, which is an anagram of &quot;ab&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length, p.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>s</code> and <code>p</code> consist of lowercase English letters.</li>
</ul>
Hash Table; String; Sliding Window
function findAnagrams(s: string, p: string): number[] {
    const m = s.length;
    const n = p.length;
    const ans: number[] = [];
    if (m < n) {
        return ans;
    }
    const cnt1: number[] = new Array(26).fill(0);
    const cnt2: number[] = new Array(26).fill(0);
    const idx = (c: string) => c.charCodeAt(0) - 'a'.charCodeAt(0);
    for (const c of p) {
        ++cnt1[idx(c)];
    }
    for (const c of s.slice(0, n - 1)) {
        ++cnt2[idx(c)];
    }
    for (let i = n - 1; i < m; ++i) {
        ++cnt2[idx(s[i])];
        if (cnt1.toString() === cnt2.toString()) {
            ans.push(i - n + 1);
        }
        --cnt2[idx(s[i - n + 1])];
    }
    return ans;
}
