<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;
        if(p.size()>s.size())
            return ans; 
        int window_size = p.size();
        int counter=0;
        map<char,int> table;
        for(int i=0;i<p.size();i++){
            if(table.find(p[i]) == table.end()){
                table.insert({p[i],1});
            }
            else{
                table[p[i]]++;
            }
        }
        map<char,int> trackerTable;
        int right=0;
        for(int i=0;i<=s.size() - window_size;i++){
            while(right<window_size){
                if(trackerTable.find(s[i+right]) == trackerTable.end()){
                    trackerTable.insert({s[i+right],1});
                }
                else{
                    trackerTable[s[i+right]]++;
                }
                right++;
            }
            if(trackerTable == table){
                ans.push_back(i);
            }
            trackerTable[s[i]]--;
            if(trackerTable[s[i]] == 0)
            {
                trackerTable.erase(s[i]);
            }
            right -=1;
        }
        return ans;
    }
};
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