<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) {
    map<char,int>mp;
    multiset<char>mt;
    vector<int>ans;
    for(int i=0;i<p.size();i++){
        mp[p[i]]++;
        mt.insert(s[i]);
    }
    //p ka size 
    //set ka maxm2 26
    //i-m ki window minus
    bool check=true;
    for(auto m:mp){
        if(m.second!=mt.count(m.first)){
            check=false;
        }
    }
    if(check){
        ans.push_back(0);
    }
    for(int i=p.size();i<s.size();i++){
        check=true;
        mt.insert(s[i]);
        mt.erase(mt.find(s[i-p.size()]));
        for(auto m:mp){
             if(m.second!=mt.count(m.first)){
             check=false;}
        }
        if(check){
            ans.push_back(i-p.size()+1);
        }
    } 
    return ans;
}
};
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