Find All Anagrams in a String

Question

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".

Tags

• Hash Table

Thought

Use the counter here to improve the efficiency and the other ideas are closed to the previous problem.

Code

from collections import Counter

class Solution(object):
    def findAnagrams(self, s, p):
        """
        :type s: str
        :type p: str
        :rtype: List[int]
        """
        result = []
        pCounter = Counter(p)
        sCounter = Counter(s[:len(p) - 1])
        for i in xrange(len(p) - 1, len(s)):
            sCounter[s[i]] += 1
            if pCounter == sCounter:
                result.append(i - len(p) + 1)
            sCounter[s[i - len(p) + 1]] -= 1
            if sCounter[s[i - len(p) + 1]] == 0:
                del sCounter[s[i - len(p) + 1]]
        return result