<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> s_anagram;
        unordered_map<char,int> p_anagram;
        for(int i=0;i<p.size();i++)
        {
            p_anagram[p[i]]++;
        }
//sliding window k concept 
//first checked fixed size window and then check remining by removing one elemmt and adding another elment
        for( int i=0;i<p.size();i++)
        {

            s_anagram[s[i]]++;
        }

        if(s_anagram==p_anagram)
        {
            ans.push_back(0);
        }

        for(int i=p.size();i<s.size();i++)
        {
            char newchar= s[i];
            char oldchar = s[i-p.size()];

             s_anagram[newchar]++;
             s_anagram[oldchar]--;
             if(s_anagram[oldchar]==0)
             {
                s_anagram.erase(oldchar);
             }


             if(s_anagram==p_anagram)
             {
                ans.push_back(i-p.size()+1);
             }

        }

        return ans;
    }
};
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