438. Find All Anagrams in a String

Problem Link

Description

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Given two strings s and p, return an array of all the start indices of p's anagrams in s. You may return the answer in any order.

 

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".

 

Constraints:

• 1 <= s.length, p.length <= 3 * 104
• s and p consist of lowercase English letters.

Solution

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Python3

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        n1, n2 = len(s), len(p)
        counter = Counter(p)
        curr = len(counter)
        res = []
        
        for i, x in enumerate(s):
            if x in counter:
                counter[x] -= 1
                if counter[x] == 0:
                    curr -= 1
            
            if curr == 0:
                res.append(i - n2 + 1)
            
            if i - n2 + 1 >= 0 and s[i - n2 + 1] in counter:
                counter[s[i - n2 + 1]] += 1
                if counter[s[i - n2 + 1]] == 1:
                    curr += 1
        
        return res