README_EN.md

438. Find All Anagrams in a String

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Description

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.

Solutions

Python3

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        counter = collections.Counter(p)
        res = []
        left = right = 0
        t = collections.Counter()
        while right < len(s):
            t[s[right]] += 1
            while t[s[right]] > counter[s[right]]:
                t[s[left]] -= 1
                left += 1
            if right - left == len(p) - 1:
                res.append(left)
            right += 1
        return res

Java

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        int[] counter = new int[26];
        for (int i = 0; i < p.length(); ++i) {
            ++counter[p.charAt(i) - 'a'];
        }
        List<Integer> res = new ArrayList<>();
        int left = 0, right = 0;
        int[] t = new int[26];
        while (right < s.length()) {
            int i = s.charAt(right) - 'a';
            ++t[i];
            while (t[i] > counter[i]) {
                --t[s.charAt(left) - 'a'];
                ++left;
            }
            if (right - left == p.length() - 1) {
                res.add(left);
            }
            ++right;
        }
        return res;
    }
}

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