# 438. Find All Anagrams in a String

{% hint style="info" %}
sliding window, (Array or HashMap)
{% endhint %}

Array

```java
class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        //sliding window: array
        int[] memo = new int[26];
        List<Integer> result = new LinkedList<Integer>();
        
        for(int i=0; i<p.length(); i++){
            memo[p.charAt(i)-'a']++;
        }
        
        int count = p.length();
        for(int i=0; i<s.length();i++){
            //first add current char to the window
            //check if memo contains curent char 
            // if current char in adjusted p string, memo >=1, count it down
            if(memo[s.charAt(i)-'a']-- >=1) count--;
             
            // then remove the first char in the window (i-p.length()),
            // if window contains enough elements. i.e. i>p.length()-1;
            // if i>p.length()-1, move the left sliding wondow
            if (i>p.length()-1 && memo[s.charAt(i-p.length())-'a']++ >=0) count++;
            
            // finally, add the first element of the window (i-p.length()+1) to result
            if(count==0) result.add(i-p.length()+1);
            
        }
        return result;
    }
}
```

HashMap

```java
class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        //sliding window: HashMap
        HashMap<Character, Integer> memo = new HashMap<Character, Integer>();
        List<Integer> result = new LinkedList<Integer>();
        
        for(int i=0; i<p.length(); i++){
            if(memo.containsKey(p.charAt(i))){
                memo.put(p.charAt(i), memo.get(p.charAt(i))+1);
            }else{
                memo.put(p.charAt(i),1);
            }
        }
        
        int count = p.length();
        for(int i=0; i<s.length();i++){
            if(memo.containsKey(s.charAt(i))){
                if (memo.get(s.charAt(i)) >= 1){
                    count--;  
                }
                memo.put(s.charAt(i), memo.get(s.charAt(i))-1);
            }
            
            if (i>p.length()-1 && memo.containsKey(s.charAt(i-p.length()))){
                if(memo.get(s.charAt(i-p.length()))>=0){       
                    count++;
                } 
                memo.put(s.charAt(i-p.length()), memo.get(s.charAt(i-p.length()))+1);
            }
            
            if(count == 0) result.add(i-p.length()+1);
        }
        return result;
    }
}
```


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