<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:
   bool check(unordered_map<char , int >&mp1, unordered_map<char  , int>&mp2){
       if(mp1.size()  != mp2.size())return false;

       for(auto it : mp2){
            if(it.second != mp1[it.first])return false;
       }
       return true;
   }
    vector<int> findAnagrams(string s, string p) {
        unordered_map<char , int>mp1; 
        unordered_map<char, int>mp2;

        for(int  i =0 ; i <p.length(); i++){
             mp1[p[i]]++;
        }

        int i=0; 
        int  j =0; 
        vector<int>ans;

        while(j< s.length()){
             if(mp1.find(s[j]) != mp1.end()){
                  mp2[s[j]]++;
             }
             j++;
             while(check(mp1, mp2)){
                cout<<i<<" "<<j<<endl;
                 ans.push_back(i);
                 if(mp1.find(s[i]) != mp1.end()){
                     mp2[s[i]]--;
                     if(mp2[s[i]] == 0){
                        mp2.erase(s[i]);
                     }
                 }
                 i++;
            }

            if(j-i+1 > p.length()){
                  if(mp1.find(s[i]) != mp1.end()){
                     mp2[s[i]]--;
                     if(mp2[s[i]] == 0){
                        mp2.erase(s[i]);
                     }
                 }
                 i++;
                 

            }
            
            

        }
        return ans;
        
    }
};
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