<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) {
        if(p.length()>s.length()) return {};
        int pl=p.length();
        map<char,int>mp;
        for(auto it:p)
        mp[it]++;
        map<char,int>mp1;
        for(int i=0;i<pl;i++)
        mp1[s[i]]++;
        int j=pl;
        int i=0;
        vector<int> ans;
        while(j<s.length())
        {
            if(mp==mp1)
            ans.push_back(i);
            mp1[s[i]]--;
            mp1[s[j]]++;
            if(mp1[s[i]]==0)
            mp1.erase(s[i]);
            i++;
            j++;
        }
        if(mp1==mp)
        ans.push_back(i);
        return ans;
    }
};
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