# 438-find-all-anagrams-in-a-string

## Question

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 pwill 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".
``````

## Thought Process

1. Map - Sliding Window
1. Use map to record the count and left and right pointers to track the windows
2. Time complexity O(n)
3. Space complexity O(n) or O(1)

## Solution

``````class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> res = new ArrayList<>();
if (s.length() < p.length()) return res;
int[] map = new int[128];
for (char c : p.toCharArray()) {
map[c]++;
}
// two pointers to track the window
int left = 0, right = 0;
char[] chars = s.toCharArray();
int m = s.length(), n = p.length(), count = n;
while (right < m) {
if (map[chars[right++]]-- > 0) count--;
if (count == 0) res.add(left);
if (right - left == n && map[chars[left++]]++ >= 0) count++;
}
return res;
}
}
``````