<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) {
        if(p.length()>s.length())
        return {};

         return checkInclusion(p,s);
        
    }
     vector<int> checkInclusion(string s1, string s2) {
        if(s2.length()<s1.length())
        return {};

        int count=0;

        map<char,int>mp;
        for(int i=0;i<s1.length();i++)
        {
            mp[s1[i]]++;
        }
        map<char,int>mp2;
       for(int i=0;i<s1.length();i++)
       {
           mp2[s2[i]]++;
           if(mp.find(s2[i])!=mp.end())
           {
              if(mp2[s2[i]]<=mp[s2[i]])
              {
                count++;
              }
           }
       }
       vector<int>ans;
       if(count==s1.length())
       {
            ans.push_back(0);
       }

       for(int i=s1.length();i<s2.length();i++)
       {
            mp2[s2[i-s1.length()]]--;
            if(mp2[s2[i-s1.length()]]==0)
            {
                mp2.erase(s2[i-s1.length()]);
            }
            if(mp.find(s2[i-s1.length()])!=mp.end())
            {
                 if(mp2[s2[i-s1.length()]]<mp[s2[i-s1.length()]])
                 {
                    count--;
                 }

            }
            mp2[s2[i]]++;

            if(mp.find(s2[i])!=mp.end())
            {
                if(mp[s2[i]]>=mp2[s2[i]])
                {
                    count++;
                }

            }

            if(count==s1.length())
           ans.push_back(i-s1.length()+1);

       }
       return ans;
        
    }
};
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