MayChallenge (47) | 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".

To solve this problem, i use the window sliding as move window folow iterator. Example: s = cbaebabacd, p = abc, futher window.length = 3. The transaction process will take place like this: [cba]ebabacd - c[bae]babacd - cb[aeb]abacd ... - cbaebab[acd]. With each every time transaction, we will check it is anagram. You can use the array frequenc to check.

Solution Find Add Anagrams in a String by Sliding Window

The solution somthing just like this:
Thank you, and see you later !