<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.size()>s.size())
        {
            return {};
        }
        s.push_back('a');
        vector<int>ans;
        int n = p.size();
        int i = 0,j= 0;
        vector<int>mp(26,0);
        for(auto i : p)
        {
            mp[i-'a']++;
        }
        p.clear();
        for(auto i : mp)
        {
            p += to_string(i);
        }
        mp.clear();
        mp.resize(26,0);
        while(i<n)
        {
            mp[s[i]-'a']++;
            i++;
        }
        string str;
        cout<<p<<endl;
        while(i<s.size())
        {
            for(auto x : mp)
            {
                str += to_string(x);
            }
            if(str == p)
            {
                ans.push_back(j);
            }
            str.clear();
            mp[s[j]-'a']--;
            mp[s[i]-'a']++;
            i++;
            j++;
        }
        return ans;
    }
};
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