438. Find All Anagrams in a String (Medium)
https://leetcode.com/problems/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".
Solutions
class Solution {
// The key idea for this problem is that anagrams are of same length and the frequence of different
// chars are identical.
// We may image a window of the same length as p and work out the character
// frequence in this window.
// We define two arrays of length 256 of which one keeps the char frequence of p and the other keeps s's.
// we update the char frequence in the window when scanning the given string. We hit the anagram string
// as long as the frequence matches.
public List<Integer> findAnagrams(String s, String p) {
List<Integer> ans = new ArrayList<>();
if (s == null || p == null || s.length() < p.length()) {
return ans;
}
// window size
int ws = p.length();
// frequence map for chars in pattern string
int[] freqP = new int[26];
Arrays.fill(freqP, 0);
for (int i = 0; i < p.length(); i++) {
freqP[p.charAt(i) - 'a']++;
}
// frequence map for chars in test string of moving window
int[] freqS = new int[26];
Arrays.fill(freqS, 0);
for (int i = 0; i < s.length(); i++) {
// The new element incorporated when window moving forward
int cRight = s.charAt(i) - 'a';
freqS[cRight]++;
if (i < ws - 1) {
continue;
}
if (i > ws - 1) {
// the leftmost element of previous window. It is edged off when window
// moving forward.
int cLeft = s.charAt(i - ws) - 'a';
freqS[cLeft]--;
}
boolean isAnagram = true;
for (int j = 0; j < 26; j++) {
if (freqS[j] == freqP[j]) {
continue;
}
isAnagram = false;
break;
}
if (isAnagram) {
ans.add(i - ws + 1);
}
}
return ans;
}
}