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
impl Solution {
    pub fn find_anagrams(s: String, p: String) -> Vec<i32> {
        let (s, p) = (s.as_bytes(), p.as_bytes());
        let (m, n) = (s.len(), p.len());
        let mut ans = vec![];
        if m < n {
            return ans;
        }

        let mut cnt = [0; 26];
        for i in 0..n {
            cnt[(p[i] - b'a') as usize] += 1;
            cnt[(s[i] - b'a') as usize] -= 1;
        }
        for i in n..m {
            if cnt.iter().all(|&v| v == 0) {
                ans.push((i - n) as i32);
            }
            cnt[(s[i] - b'a') as usize] -= 1;
            cnt[(s[i - n] - b'a') as usize] += 1;
        }
        if cnt.iter().all(|&v| v == 0) {
            ans.push((m - n) as i32);
        }
        ans
    }
}
