<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) {

        // vector<int> ans ;
        // unordered_map <char, int> mp1, mp2 ;

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

        // int i = 0 ;
        // int j = 0 ;
        // int n = s.size() ;
        // int m = p.size() ; 

        // while(j < n){

        //     mp2[s[j]]++ ; 

        //     if(j-i+1 < m){
        //         j++ ; 
        //     }

        //     else if(j-i+1 == m){

        //         if(mp1 == mp2){
        //             ans.push_back(i) ;
        //         }

        //         mp2[s[i]] -- ;
        //         if(mp2[s[i]] == 0){
        //             mp2.erase(s[i]) ; 
        //         }

        //         i++ ;
        //         j++ ; 
        //     }
        // }

        // return ans ; 


        vector<int> ans ;
        unordered_map<char, int> mp ;

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

        int i = 0 ;
        int j = 0 ;
        int n = s.size() ;
        int m = p.size() ; 
        int count = mp.size() ;

        while(j < n){

            if(mp.find(s[j]) != mp.end()){

                mp[s[j]]-- ; 

                if(mp[s[j]] == 0){
                    count-- ;

                    if(count == 0){
                        ans.push_back(i) ; 
                    }
                }
            }

            if(j-i+1 < m){
                j++ ; 
            }

            else if(j-i+1 == m){

                // slide the window
                if(mp.find(s[i]) != mp.end()){

                    mp[s[i]]++ ;

                    if(mp[s[i]] == 1){
                        count++ ; 
                    }
                }

                i++ ;
                j++ ; 
            }
        }

        return ans; 
        
    }
};
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