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438. Find All Anagrams in a String

2024-10-03 18:49:02   212人阅读

https://leetcode.com/problems/find-all-anagrams-in-a-string/#/description

 

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".

 

 

Sol:

 

 

class Solution(object):
    def findAnagrams(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: List[int]
        """
        from collections import Counter

        res = []
        pCounter = Counter(p)
        sCounter = Counter(s[:len(p)-1])
        for i in range(len(p)-1,len(s)):
            sCounter[s[i]] += 1   # include a new char in the window
            if sCounter == pCounter:    # This step is O(1), since there are at most 26 English letters 
                res.append(i-len(p)+1)   # append the starting index
            sCounter[s[i-len(p)+1]] -= 1   # decrease the count of oldest char in the window
            if sCounter[s[i-len(p)+1]] == 0:
                del sCounter[s[i-len(p)+1]]   # remove the count if it is 0
        return res

 

 

 

Note:

 

1 counter usage: 

print collections.Counter([‘a‘, ‘b‘, ‘c‘, ‘a‘, ‘b‘, ‘b‘])

print collections.Counter({‘a‘:2, ‘b‘:3, ‘c‘:1})
print collections.Counter(a=2, b=3, c=1)
 
 
 
==>
Counter({‘b‘:3, ‘a‘: 2, ‘c‘: 1})
Counter({‘b‘:3, ‘a‘: 2, ‘c‘: 1})
Counter({‘b‘:3, ‘a‘: 2, ‘c‘: 1})

438. Find All Anagrams in a String


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