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".```

Python

```
class Solution(object):
def findAnagrams(self, s, p):
"""
:type s: str
:type p: str
:rtype: List[int]
"""
result = []
if not s:
return []
p_dict = dict()
for _p in p:
s_dict = dict()
index = 0
for _s in s[index: index + len(p)]:
if s_dict == p_dict:
result.append(index)
index += 1
while index+len(p) <= len(s):
print index+len(p)
Solution.del_k_from_dict(s[index-1], s_dict)
if s_dict == p_dict:
result.append(index)
index += 1
return result

@classmethod