438. Find All Anagrams in a String

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.

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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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class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        int[] freq = new int[26];
        int[] sfreq = new int[26];
        int len = p.length();
        List<Integer> res = new ArrayList<>();
        if (len > s.length()) {
            return res;
        }
        for (int i = 0; i < len; i++) {
            sfreq[s.charAt(i) - 'a']++;
            freq[p.charAt(i) - 'a']++;
        }
        if (Arrays.equals(freq, sfreq)) {
            res.add(0);
        }
        for (int i = len; i < s.length(); i++) {
            sfreq[s.charAt(i) - 'a']++;
            sfreq[s.charAt(i - len) - 'a']--;
            if (Arrays.equals(freq, sfreq)) {
                res.add(i - len + 1);
            }
        }
        return res;
    }
}

Solution#

Solution