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
public class Solution {
    public IList<int> FindAnagrams(string s, string p) {
        int m = s.Length, n = p.Length;
        IList<int> ans = new List<int>();
        if (m < n) {
            return ans;
        }
        int[] cnt1 = new int[26];
        int[] cnt2 = new int[26];
        for (int i = 0; i < n; ++i) {
            ++cnt1[p[i] - 'a'];
        }
        for (int i = 0, j = 0; i < m; ++i) {
            int k = s[i] - 'a';
            ++cnt2[k];
            while (cnt2[k] > cnt1[k]) {
                --cnt2[s[j++] - 'a'];
            }
            if (i - j + 1 == n) {
                ans.Add(j);
            }
        }
        return ans;
    }
}
