Given two strings s and p, return an array of all the start indices ofp's anagrams ins. You may return the answer in any order.
An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
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".
Constraints:
1 <= s.length, p.length <= 3 * 104
s and p consist of lowercase English letters.
ac
class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> res = new ArrayList<>();
//edge cases
if (s.length() < p.length() || s.length() == 0) return res;
// get char counts
int[] map = new int[26];
int diff = 0;
for (char c : p.toCharArray()) {
map[c-'a']++;
if (map[c-'a'] == 1) diff++;
}
// sliding window
int l = 0;
for (int r = 0; r < s.length(); r++) {
char c = s.charAt(r);
map[c-'a']--;
if (map[c-'a'] == 0) diff--;
else if (map[c-'a'] == -1) diff++;
if (r - l >= p.length()) {
char lc = s.charAt(l++); // move left point
map[lc-'a']++;
if (map[lc-'a'] == 0) diff--;
else if (map[lc-'a'] == 1) diff++;
}
if (diff == 0) res.add(l);
}
return res;
}
}
// same as https://leetcode.com/problems/permutation-in-string/description/