<p>Given two strings <code>s</code> and <code>p</code>, return <em>an array of all the start indices of </em><code>p</code><em>&#39;s anagrams in </em><code>s</code>. You may return the answer in <strong>any order</strong>.</p>

<p>An <strong>Anagram</strong> is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.</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>

class Solution {
public:
  vector<int> findAnagrams(string s, string p) {
    int m = s.length(), n = p.length();
    if (m < n) return vector<int>();
    
    map<char, int> cnt;
    for (char ch : p) --cnt[ch];
    
    for (int i = 0; i < n-1; ++i) {
      char ch = s[i];
      ++cnt[ch];
      if (cnt[ch] == 0) cnt.erase(ch);
    }
    
    vector<int> res;
    
    for (int i = n-1; i < m; ++i) {
      char ch = s[i];
      ++cnt[ch];
      if (cnt[ch] == 0) cnt.erase(ch);
      
      if (cnt.empty()) {
        res.push_back(i - n + 1);
      }
      
      // Remove(cnt, s[i-n+1])
      ch = s[i-n+1];
      --cnt[ch];
      if (cnt[ch] == 0) cnt.erase(ch);
    }
    
    return res;
  }
};
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