<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> search(string pat, string txt) {
	    // code here
	    map<char,int>mp; vector<int> ans;
	    //char,frew
	    for(auto it:pat)
	    {
	        mp[it]++;
	    }
	    int k=pat.size();
	    int i=0;
	    int j=0; int count=mp.size();
	    while(j<txt.size())
	    {
	        if(mp.find(txt[j])!=mp.end())
	        {
	            mp[txt[j]]--;
	            if(mp[txt[j]]==0){
	               // mp.erase(mp[txt[j]]); 
	            count--;}
	            
	        }
	        if(j-i+1<k)j++;
	        else if(j-i+1==k)
	        {
	            if(count==0){
	                ans.push_back(i);
	                
	            }
	           // mp[txt[i]]++;
	           //     if(mp[txt[i]]==1)count++;
	           //    // mp[txt[i]]++; count++;
	           //    // if(mp[txt[i]]==0)mp.erase(mp[txt[i]]); 
	           //    //count=mp.size();
	            if (mp.find(txt[i]) != mp.end()) {
                mp[txt[i]]++;
                if (mp[txt[i]] == 1) {
                    count++; // If frequency becomes positive, it means we are missing this character in the window
                }
            }
	               i++;j++;
	            }
	            
	        }
	     return ans;
	    }
	
    vector<int> findAnagrams(string s, string p) {
        return search(p,s);
    }
};
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