<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) {
        int n = s.size();
        
        map<char, int> mp;
        int k = p.size();
        for(int i = 0; i < k; i++){
            mp[p[i]]++;
        }
        
        int count = mp.size();
        
        int i = 0;
        int j = 0;
       
        vector<int> v;
        
        while(j < n){
            if(mp.count(s[j]) == 1){
                mp[s[j]]--;
                
                if(mp[s[j]] == 0){
                    count--;
                }
            }
            
            int l = (j - i + 1);
            
            if(l < k){
                j++;
            }
            else if(l == k){
                if(count == 0){
                    v.push_back(i);
                }
                
                if(mp.count(s[i]) == 1){
                    mp[s[i]]++;
                    if(mp[s[i]] == 1){
                        count++;
                    } 
                }
               
                i++;
                j++;
            }
        }
        
        return v;
    }
};
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