Find All Anagrams in a String

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".

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> list = new ArrayList<>();
        if (s == null || s.length() == 0 || p == null || p.length() == 0)
            return list;
        int[] hash = new int[256];
        for (char c : p.toCharArray())
            hash[c]++;

        int left = 0, right = 0, count = p.length();
        while (right < s.length()) {
            // Include first element
            char c = s.charAt(right);
            if (hash[c] > 0)
                count--;
            hash[c]--;
            // Maintaining a sliding window of size -> p.size()
            if (right - left + 1 == p.length()) {
                char c1 = s.charAt(left);
                if (count == 0)
                    list.add(left);
                if (hash[c1] >= 0)
                    count++;
                hash[c1]++;
                // Kick last element, for the next window
                left++;
            }
            right++;
        }
        return list;
    }
}