<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>

from collections import Counter

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        sLen = len(s)
        pLen = len(p)
        if pLen > sLen:
            return []
        
        # results = []
        # window = s[:pLen]
        # sCtr = Counter(window)
        # pCtr = Counter(p)

        # for i in range(sLen):
        #     # print(window)
        #     checkFlag = True
        #     for char in window:
        #         if (char not in pCtr) or (sCtr[char] != pCtr[char]):
        #             checkFlag = False
        #             break
        #     if checkFlag:
        #         results.append(i)
        #     if i + pLen >= sLen:
        #         break

        #     sCtr[window[0]] -= 1
        #     window = window[1:]

        #     window += s[i + pLen]
        #     if window[-1] not in sCtr:
        #         sCtr[window[-1]] = 0
        #     sCtr[window[-1]] += 1
        
        # return results
            
        results = []
        counts = [0] * 26

        for c in p:
            counts[ord(c) - ord('a')] += 1
        
        left, right = 0, 0
        while right < len(s):
            counts[ord(s[right]) - ord('a')] -= 1
            while left <= right and counts[ord(s[right]) - ord('a')] < 0:
                counts[ord(s[left]) - ord('a')] += 1
                left += 1
            if right - left + 1 == len(p):
                results.append(left)
            right += 1
        
        return results
            

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