Leetcode: 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.

Sliding Window: Use a count to denote the difference between current sliding window and p, if count == 0, means the current sliding window is the same with p, add start to the result

refer to https://discuss.leetcode.com/topic/64434/shortest-concise-java-o-n-sliding-window-solution/2

public 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]; //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) { //char at right is in p, so is needed to make an anagram of p, so if we shift our window to
            count--;　　　　　　　　　　　　　// include right, the difference between our window and p will decrease by 1, so count-1
        }
        hash[s.charAt(right)]--;　　　　　　//Other case if hash[s.charAt(right)] right now == 0, which means we do not need it, we do not decrease count
        right++;　　　　　　　　　　　　　　　　// hash[s.charAt(right)] will be -1, if window contains some char that p do not need, count will never reach 0
        　　　　　　　　　　　　　　　　　　　　　// whereas some hash[somewhere] will be negative, some hash[somewhereelse] will be positive
        //when the count is down to 0, means we found the right anagram
        //then add window's left to result list
        if (count == 0) {　　　　　　　　　　// this is only when p's chars and window's chars are all the same
            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() ) {
           
            if (hash[s.charAt(left)] >= 0) {
                count++;
            }
            hash[s.charAt(left)]++;
            left++;
        
        }

        
    }
        return list;
    }
}