> For the complete documentation index, see [llms.txt](https://jaywin.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://jaywin.gitbook.io/leetcode/solutions/0438-find-all-anagrams-in-a-string.md).

# 0438. Find All Anagrams in a String

<https://leetcode.com/problems/find-all-anagrams-in-a-string>

## 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**.

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.

## ac

```java
class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> res = new ArrayList<>();
        //edge cases
        if (s.length() < p.length() || s.length() == 0) return res;

        // get char counts
        int[] map = new int[26];
        int diff = 0;
        for (char c : p.toCharArray()) {
            map[c-'a']++;
            if (map[c-'a'] == 1) diff++;
        }

        // sliding window
        int l = 0;
        for (int r = 0; r < s.length(); r++) {
            char c = s.charAt(r);
            map[c-'a']--;
            if (map[c-'a'] == 0) diff--;
            else if (map[c-'a'] == -1) diff++;

            if (r - l >= p.length()) {
                char lc = s.charAt(l++); // move left point
                map[lc-'a']++;
                if (map[lc-'a'] == 0) diff--;
                else if (map[lc-'a'] == 1) diff++;
            }

            if (diff == 0) res.add(l);
        }

        return res;
    }
}

// same as https://leetcode.com/problems/permutation-in-string/description/
```
