438. Find All Anagrams in a String

Description

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".

Solution

Two-pointer

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> anagrams = new ArrayList<>();
        if (s == null || p == null || s.isEmpty() || p.isEmpty() || s.length() < p.length()) {
            return anagrams;
        }
        
        int[] need = new int[256];
        
        for (char c : p.toCharArray()) {
            ++need[c];
        }
        
        int left = 0;
        int right = 0;
        int stillNeed = 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 (need[s.charAt(right)] > 0) {
                --stillNeed;
            }
            --need[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 (stillNeed == 0) {
                anagrams.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 (need[s.charAt(left)] >= 0) {
                    ++stillNeed;
                }
                
                ++need[s.charAt(left)];
                ++left;
            }
        }
        
        return anagrams;
    }
}