<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_map<char,int> mp;

        unordered_map<char,set<int>> mp1;
        for(auto it:p){
            mp[it]++;
        }

        int i=0,j=0;
        int n=s.size();
         int cnt=0;
         int m=p.size();
         vector<int> ans;
        
        while(i<=j and j<n){
            if(mp.find(s[j])!=mp.end()){
                if(mp1[s[j]].size()<mp[s[j]]){
                    mp1[s[j]].insert(j);
                    j++;
                    cnt++;
                    if(cnt==m){
                        ans.push_back(i);
                    }
                }else{
                   
                   while(s[i]!=s[j]){
                    mp1[s[i]].erase(mp1[s[i]].begin());
                    cnt--;
                    i++;
                   }
                   mp1[s[i]].erase(mp1[s[i]].begin());
                   cnt--;
                   i++;
                   
                }
            }else{
               
                   cnt=0;
                    j++;
                    i=j;
                    mp1.clear();
                
            }
        }
        

        return ans;
    }
};
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