<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:
    bool checkAna(vector<int> &a, vector<int> b) {
        for(int i = 0; i < 26; i++) {
            if (a[i] != b[i]) return false;
        }
        return true;
    }
    vector<int> findAnagrams(string s, string p) {
        if(p.size() > s.size())
        {
            return {};
        }
        vector<int> mpS(26, 0);
        vector<int> mpP(26, 0);
        int n = s.length();
        int m = p.length();
        vector<int> soln;
        for(auto &x: p) {
            mpP[x - 'a']++;
        }
        int st = 0, ed = 0;
        while(ed < n) {
            mpS[s[ed] -'a']++;
            if(ed -st + 1 == m) {
                if(checkAna(mpS, mpP)) soln.push_back(st);
            }
            if(ed - st + 1>= m) {
                mpS[s[st] - 'a']--;
                st++;
            }
            ed++;
        }
        return soln;
    }
};

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