# 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".
# Solution
Approach 1: sliding window.
# Code (Python)
Approach 1:
class Solution:
def findAnagrams(self, s: str, p: str) -> List[int]:
result = []
freqs_in_p = [0] * 26
for letter in p:
freqs_in_p[ord(letter) - ord('a')] += 1
freqs_in_p = tuple(freqs_in_p)
freqs = [0] * 26
for r in range(len(s)):
freqs[ord(s[r]) - ord('a')] += 1
if r >= len(p):
freqs[ord(s[r-len(p)]) - ord('a')] -= 1
if tuple(freqs) == freqs_in_p:
result.append(r - len(p) + 1)
return result
# Code (C++)
Approach 1:
Approach 2: