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".
class Solution {
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];
for (char c : p.toCharArray())
hash[c]++;
int left = 0, right = 0, count = p.length();
while (right < s.length()) {
// Include first element
char c = s.charAt(right);
if (hash[c] > 0)
count--;
hash[c]--;
// Maintaining a sliding window of size -> p.size()
if (right - left + 1 == p.length()) {
char c1 = s.charAt(left);
if (count == 0)
list.add(left);
if (hash[c1] >= 0)
count++;
hash[c1]++;
// Kick last element, for the next window
left++;
}
right++;
}
return list;
}
}