438. Find All Anagrams in a String

438. Find All Anagrams in a String

Medium

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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.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

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.

Accepted

507,570

Submissions

1,056,733

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        dict_p,dict_s=Counter(p),defaultdict(int)
        num_to_correct,ans=len(dict_p),[]
        
        for i in range(len(s)):
            new_s=s[i]
            dict_s[new_s]+=1
            if dict_s[new_s]==dict_p[new_s]:
                num_to_correct-=1
            if dict_s[new_s]==dict_p[new_s]+1:
                num_to_correct+=1
                
            if i >= len(p):
                old_s=s[i-len(p)]
                dict_s[old_s]-=1
                if dict_s[old_s]==dict_p[old_s]-1:
                    num_to_correct+=1
                if dict_s[old_s]==dict_p[old_s]:
                    num_to_correct-=1
            
            if num_to_correct==0: ans.append(i+1-len(p))
        return ans

dict_s Record each char The number of

Every time the quantity changes , Maintenance