<p>Given two strings <code>s</code> and <code>p</code>, return an array of all the start indices of <code>p</code>&#39;s <span data-keyword="anagram">anagrams</span> in <code>s</code>. You may return the answer in <strong>any order</strong>.</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:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        alpha2int = {c:i for i,c in enumerate("abcdefghijklmnopqrstuvwxyz")}
        p_char_count = [0]*26
        s_char_count = [0]*26
        ans = []
        n = len(p)

        for c in p:
            p_char_count[alpha2int[c]] += 1

        for idx, c in enumerate(s):
            s_char_count[alpha2int[c]] += 1
            if idx<n-1:
                continue
            if idx>=n:
                s_char_count[alpha2int[s[idx-n]]] -=1
            
            is_anagram = True
            for alpha_idx in range(26):
                if p_char_count[alpha_idx]!=s_char_count[alpha_idx]:
                    is_anagram = False
                    break
            
            if is_anagram:
                ans.append(idx-n+1)
        return ans
        
17700