<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:

/*
  Solution :

  1.Sliding Window  approach.

  2. Store frequency of charcter in array or map.

  3. s = "cbaebabacd", p = "abc".
   here sliding window of size=3;

       i] cbaebabacd    (cba) i=0;  ans
          ---
       ii] cbaebabacd   (bae) i=1;
            ---
       iii] cbaebabacd   (aeb) i=2;
              ---
        iv] cbaebabacd    (bab) i=3;
                ---   
        v] cbaebabacd     (aba) i=4;
                ---   
        vi] cbaebabacd    (bac) i=5;  ans
                  ---   
        vii] cbaebabacd   (acd) i=6;
                    ---  

*/
    vector<int> findAnagrams(string s, string p) {
    vector<int>ans;
    map<int,int>mp1,mp2;
    for(auto x:p)  // Store frequency of charcters of p in mp1
    {
    mp1[x]++;
    }

    int i=0;
    int j=p.size();
    int pointer=i;
    while(pointer<j)   // Pre-compute first sliding window 
    {
        mp2[s[pointer]]++;
        pointer++;
    }

    while(j<s.size())   // Move slide 
    {
        if(mp1==mp2)
        {
            ans.push_back(i);
        }

        mp2[s[i]]--;
        mp2[s[j]]++;

        if(mp2[s[i]]==0)
        {
            mp2.erase(s[i]);
        }
        i++;
        j++;

    }

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


   return ans;

        
    }
};
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