Find All Anagrams in a String

438. 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".

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;        
        }
        
        for (int i = 0; i <= s.length() - p.length();i++) {
            String str = s.substring(i, i + p.length());
            
            if (isAnagram(str, p)) {
                list.add(i);
            }
        }
        
        return list;
    }
    
    private boolean isAnagram(String s, String t) {
        if (s == null || t == null) {
            return false;
        }
        
        if (s.length() != t.length()) {
            return false;
        }
        
        int[] charCount = new int[256];
        for (int i = 0; i < s.length(); i++) {
            charCount[s.charAt(i)]++;
        }
        
        for (int i = 0; i < t.length(); i++) {
            if (--charCount[t.charAt(i)] < 0) {
                return false;
            }
        }

        return true;
    }
}

public class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> result = new ArrayList<>();
        if (s == null || s.length() == 0
            || p == null || p.length() == 0) {
            return result;        
        }
        
        // record each character in p
        int[] charCount = new int[256];
        for (char c : p.toCharArray()) {
            charCount[c]++;
        }
        
        int left = 0, right = 0; // two pointers
        int count = p.length(); // initialize count to p's 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 (charCount[s.charAt(right)] >= 1) {
                count--;
            }
            
            charCount[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 (count == 0) {
                result.add(left);
            }
            
            // if we find the window's size equals to p, then we have to move left 
            // 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 (charCount[s.charAt(left)] >= 0) {
                    count++;
                }
                
                charCount[s.charAt(left)]++;
                left++;
            }
        }
        
        return result;
    }
}

Hope this helps,
Michael