Java – HashMap containskey complexity

I have a method. I write a duplicate item in the list It works well, but I'm worried about the complexity of using containskey When we use containskey, we have to calculate a hash function for each key and compare each key with our search term, right? Isn't that O (n)?

This is the function:

public void findDup(List<String> list){

    HashMap<String,Integer> map = new HashMap<>();
    int pos=0;
    for(String s: list){
        if(map.containsKey(s)){
            Log.v("myapp","duplicate found:"+s);
        }
        else
            map.put(s,pos);
        pos++;
    }
}

And call me doing this:

List<String>list=new ArrayList<>();

    for(int i=0;i<12;i++)
        list.add(i+"");

    //these numbers should surely be duplicates
    list.add("3");list.add("6");

    findDup(list);

//The output will be clearly displayed as 3 and 6

Update: I rewritten the function, using only a more meaningful set:

public void findDup(List<Integer> list){

        HashSet<Integer> set = new HashSet<>();
        for(Integer num: list){
            if(!set.add(num)){
                Log.v("myapp","duplicate found:"+num);
            }

        }
    }

Solution

It is specified as O (1) in Javadoc

Therefore, the complexity of the algorithm is O (n)

But even if there is no containskey () call, this is actually unnecessary All you have to do is test whether put () returns a non - null value, indicating a duplicate

Wrong We calculate the hash value of the search keyword and check whether the bucket is occupied by equal keys

No,