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".
Time: O(N)
class Solution: def findAnagrams(self, s: str, p: str) -> List[int]: my_dict = {} res = [] count = 0 for char in p: freq = my_dict.get(char, 0) my_dict[char] = freq + 1 for i in range(len(s)): char = s[i] if char in my_dict: my_dict[char] -= 1 if my_dict[char] == 0: count += 1 if i >= len(p): start = i - len(p) start_char = s[start] if start_char in my_dict: my_dict[start_char] += 1 if my_dict[start_char] == 1: count -= 1 # check my_dict size instead of len(p) if count == len(my_dict): res.append(i - len(p) + 1) return res
public class Solution { public List<Integer> allAnagrams(String sh, String lo) { // Write your solution here Map<Character, Integer> map = new HashMap<>(); char[] charArr = sh.toCharArray(); for (int i = 0; i < charArr.length; i++) { map.put(charArr[i], map.getOrDefault(charArr[i], 0) + 1); } List<Integer> res = new ArrayList<>(); char[] lCharArr = lo.toCharArray(); int count = 0; int start = 0; for (int i = 0; i < lCharArr.length; i++) { char cur = lCharArr[i]; if (map.containsKey(cur)) { int num = map.get(cur); if (num == 1) { count += 1; } map.put(cur, num - 1); } if (i >= sh.length()) { start = i - sh.length(); if (map.containsKey(lCharArr[start])) { int startNum = map.get(lCharArr[start]); if (startNum == 0) { count -= 1; } map.put(lCharArr[start], startNum + 1); } } if (count == map.size()) { res.add(i - sh.length() + 1); } } return res; } }