<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:
    std::vector<int> findAnagrams(std::string s, std::string p) {
        std::unordered_map<char, std::pair<int, int>> um;
        for (auto &c : p) {
            um[c] = {um[c].second+1, um[c].second+1};
        }
        
        std::vector<int> res;
        std::queue<char> q;
        for (int i = s.size()-1; i >= 0; --i) {
            if (!um[s[i]].first) {
                q = std::queue<char>();
                for (auto & [key, val] : um) {
                    um[key].second = val.first;
                }
            } else {
                while (!um[s[i]].second && !q.empty()) {
                    char c = q.front(); q.pop();
                    ++um[c].second;
                    if (c == s[i]) break;
                }
                q.push(s[i]);
                --um[s[i]].second;
                if (q.size() == p.size()) res.push_back(i);
            }
            
        }
        
        return res;
    }
};
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