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".

class Solution(object):
    def findAnagrams(self, s, p):
        
        lp = len(p)
        ls = len(s)
        
        ds = {}
        dp = {}
        
        
        for i in s[:lp]:
            ds[i] = ds.get(i,0) + 1
        for i in p[:lp]:
            dp[i] = dp.get(i,0) + 1
            
        
        res = []
        i = 0
        
        while i < (ls-lp):
            if dp == ds:
                res += i,
            
            ds[s[i]] -= 1
            if ds[s[i]] == 0:
                del ds[s[i]]
            ds[s[i+lp]] = ds.get(s[i+lp],0) + 1
            i+= 1
        if ds == dp:
            res += i,
            
        return res
        """
        :type s: str
        :type p: str
        :rtype: List[int]
        """