[Leetcode/438] Find All Anagrams in a String #slidingwindow

Leetcode_438_Solution_A.java

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> result = new ArrayList<>();
        if (s == null || s.length() == 0) {
            return result;
        }
        if (p == null || p.length() > s.length()) {
            return result;
        }
        int[] s_freq = new int[26];
        int[] p_freq = new int[26];
        for (int i = 0; i < p.length(); i++) {
            s_freq[s.charAt(i) - 'a']++;
            p_freq[p.charAt(i) - 'a']++;
        }
        int left = 0;
        int right = p.length();
        if (Arrays.equals(s_freq, p_freq)) {
            result.add(left);
        }
        while (right < s.length()) {
            s_freq[s.charAt(left++) - 'a']--;
            s_freq[s.charAt(right++)- 'a']++;
            if (Arrays.equals(s_freq, p_freq)) {
                result.add(left);
            }
        }
        return result;
    }
}

Leetcode_438_Solution_A.py

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        result = []
        if len(s) < len(p):
            return result

        s_freq = [0 for _ in range(26)]
        p_freq = [0 for _ in range(26)]

        for i in range(len(p)):
            s_freq[ord(s[i]) - ord('a')] += 1
            p_freq[ord(p[i]) - ord('a')] += 1
        
        if s_freq == p_freq:
            result.append(0)

        for i in range(len(p), len(s)):
            s_freq[ord(s[i - len(p)]) - ord('a')] -= 1
            s_freq[ord(s[i]) - ord('a')] += 1
            if s_freq == p_freq:
                result.append(i - len(p) + 1)
        return result