<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]:
        l = 0
        result = []
        p_char_map = {}
        s_char_map = {}
        for c in p:
            p_char_map[c] = p_char_map.get(c, 0) + 1

        for r in range(len(s)):
            s_char_map[s[r]] = s_char_map.get(s[r], 0) + 1

            if s_char_map == p_char_map:
                result.append(l)

            if sum(p_char_map.values()) == sum(s_char_map.values()):
                s_char_map[s[l]] -= 1
                if s_char_map[s[l]] == 0:
                    del s_char_map[s[l]]
                l += 1

        return result
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