438. 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".
Time Complexity will be O(n) because the "start" and "end" points will only move from left to right once.
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]; //character hash //record each character in p to hash for (char c : p.toCharArray()) { hash[c]++; } //two points, initialize count to p's length int left = 0, right = 0, count = p.length(); while (right < s.length()) { //move right everytime, if the character exists in p's hash, decrease the count //current hash value >= 1 means the character is existing in p if (hash[s.charAt(right)] >= 1) { count--; } hash[s.charAt(right)]--; right++; //when the count is down to 0, means we found the right anagram //then add window's left to result list if (count == 0) list.add(left); //if we find the window's size equals to p, then we have to move left (narrow the window) to find the new match window //++ to reset the hash because we kicked out the left //only increase the count if the character is in p //the count >= 0 indicate it was original in the hash, cuz it won't go below 0 if (right - left == p.length() && hash[s.charAt(left)] >= 0) { count++; } if (right - left == p.length()) { hash[s.charAt(left)]++; left++; } } return list; }