<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>mp1(26,0);
        vector<int>mp2(26,0);
        vector<int>result;
        int hash1=0,hash2=0,st=-1,j;
        for(auto x:p)
        {
             
             if(mp1[x-'a']==0)
             hash1=hash1+pow(2,x-'a');
             mp1[x-'a']++;
             
        }
        for(int i=0;i<s.size();i++)
        {
            if(mp2[s[i]-'a']==0)
            hash2=hash2+pow(2,s[i]-'a');
            mp2[s[i]-'a']++;
            if(i<p.size()-1)
            continue;
            st++;
            if(hash1==hash2)
            {
                for(j=0;j<26;j++)
                {
                    if(mp1[j]!=mp2[j])
                    break;
                }
                if(j==26)
                result.push_back(st);
            }
            mp2[s[st]-'a']--;
            if(mp2[s[st]-'a']==0)
            hash2=hash2-pow(2,s[st]-'a');

        }
        return result;
    }
};
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