<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) {
        list<char> anagram;
        vector<int> result;
        
        unordered_map<char, int> p_ft;
        unordered_map<char, int> anagram_ft;
        
        for(auto letter: p)
            p_ft[letter]++;

        for(int i = 0; i < s.size(); i++){
            anagram.push_back(s[i]);
            anagram_ft[s[i]]++;

            if(anagram.size() < p.size())
                continue;
            
            if(anagram.size() == p.size()){
                if(is_anagram(anagram_ft, p_ft))
                    result.push_back(i - p.size() + 1);
            } else {
                auto first_letter = anagram.front();
                anagram.pop_front();
                anagram_ft[first_letter]--;
                if(is_anagram(anagram_ft, p_ft))
                    result.push_back(i - p.size() + 1);
            }
        }

        return result;
    }

    bool is_anagram(unordered_map<char, int>& anagram_ft, unordered_map<char, int>& p_ft){
        for(auto&[letter, frequency]: p_ft)
            if(anagram_ft.find(letter) == anagram_ft.end() || frequency != anagram_ft[letter])
                return false;
        
        return true;
    }
};
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