# Leetcode: Find All Anagrams in a String

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".
```

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```## Blog link: https://code.dennyzhang.com/find-all-anagrams-in-a-string
class Solution(object):
def is_same_dict(self, dict_1, dict_2):
has_matched = True
if len(dict_1.keys()) == len(dict_2.keys()):
for k in dict_1:
if dict_1[k] != dict_2[k]:
has_matched = False
break
else:
has_matched = False

return has_matched

def findAnagrams(self, s, p):
"""
:type s: str
:type p: str
:rtype: List[int]
"""
p_dict = {}
q_dict = {}
ret = []

if len(s) < len(p):
return []

for ch in 'abcdefghijklmnopqrstuvwxyz':
p_dict[ch] = 0
for ch in p:
p_dict[ch] += 1

for ch in 'abcdefghijklmnopqrstuvwxyz':
q_dict[ch] = 0
for ch in s[0:len(p)]:
q_dict[ch] += 1

if self.is_same_dict(p_dict, q_dict) is True:
ret.append(0)

# loop the change
for i in range(1, len(s)-len(p)+1):
# print q_dict
q_dict[s[i+len(p)-1]] += 1
q_dict[s[i-1]] -= 1
if self.is_same_dict(p_dict, q_dict) is True:
ret.append(i)

return ret
```

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