<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) {
        // unordered_set<string> st;
        // vector<int> res;
        // if(s.size()<p.size())
        //     return res;
        // sort(p.begin(),p.end());
        // st.insert(p);
        // int sw = p.size();
        
        // for(int i=0;i<s.size()-sw+1;i++) {
        //     string temp = s.substr(i,sw);
        //     sort(temp.begin(),temp.end());
        //     if(st.count(temp))
        //         res.push_back(i);
        // }
        // return res;
        vector<int> res;
        if(s.size()<p.size())
             return res;
        map<char,int> mp;
        for(auto& c:p)
            mp[c]++;
        for(int i=0;i<p.size();i++) {
            mp[s[i]]--;
            if(mp[s[i]]==0)
                mp.erase(s[i]);
        }
        if(mp.size()==0)
             res.push_back(0);
        int sw = p.size();
        for(int i=sw;i<s.size();i++) {
            mp[s[i-sw]]++;
            mp[s[i]]--;
            if(mp[s[i]]==0)
                mp.erase(s[i]);
            if(mp[s[i-sw]]==0)
                mp.erase(s[i-sw]);
            if(mp.size()==0)
                res.push_back(i-sw+1);
        }
        return res;
    }
};
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