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".

思路：可以用sliding window，也可以用two pointer
two pointer每步都要决定怎么更新pointer，sliding window就是有些先验知识，可以不必每步更新，需要的时候跳到另外一个地方

# two pointer: decision at every step for pointers
# sliding window, each step, point may not change, but can jump to another place
from collections import defaultdict, Counter
class Solution:
def findAnagrams(self, s, p):
"""
:type s: str
:type p: str
:rtype: List[int]
"""
counts = dict(Counter(p))
runningDict = defaultdict(int)
res = []
i, j = 0, 0
while j<len(s):
# choose how to change i,j point at every step
if s[j] not in counts:
j += 1
i = j
runningDict.clear()
elif runningDict[s[j]] < counts[s[j]]:
runningDict[s[j]] += 1
j += 1
elif runningDict[s[j]] == counts[s[j]]:
runningDict[s[i]] -= 1
i += 1

if counts == runningDict:
res.append(i)
return res

s=Solution()
print(s.findAnagrams('cbaebabacd', 'abc'))
print(s.findAnagrams('abab', 'ab'))