438. Find All Anagrams in a String

Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:
s: "cbaebabacd" p: "abc"

Output:
[0, 6]

Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc". 

Example 2:

Input:
s: "abab" p: "ab"

Output:
[0, 1, 2]

Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".
分析：
这题和mininum window几乎一摸一样，使用两个pointer来找这样的window。

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        from collections import defaultdict
        letter_count = defaultdict(int)
        start = end = 0
        result = []
        for letter in p:
            letter_count[letter] += 1
        count = len(letter_count)
        while end < len(s):
            if s[end] in letter_count:
                letter_count[s[end]] -= 1
                if letter_count[s[end]] == 0:
                    count -= 1
            end += 1
            while count == 0 and start < len(s):
                if end - start == len(p):
                    result.append(start)
                if s[start] in letter_count:
                    letter_count[s[start]] += 1
                    if letter_count[s[start]] == 1:
                        count += 1
                start += 1
        return result