<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 l=0,r=0,count=0,slen=s.length(),plen=p.length();
        vector<int> res;
        
        unordered_map<char,int> ref,anagram;

        if (plen > slen) return res;
        
        for (auto i: p) ref[i]++;

        while (r < slen){
            if (ref.find(s[r]) != ref.end()){
                anagram[s[r]]++;
                count++;
                while (l<r && anagram[s[r]] > ref[s[r]]){
                    if (ref.find(s[l]) != ref.end()){
                        anagram[s[l]]--;
                        count--;
                    }
                    l++;
                }
                if (count == plen){
                    res.push_back(l);
                    anagram[s[l]]--;
                    l++;
                    count--;
                }
            }else{
                anagram.clear();
                count = 0;
                l = r+1;
            }
            r++;
        }
        return res;
    }
};
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