Distinct Numbers in Window

You are given an array of N integers, A1, A2 ,…, AN and an integer K. Return the of count of distinct numbers in all windows of size K.

Formally, return an array of size N-K+1 where i’th element in this array contains number of distinct elements in sequence Ai, Ai+1 ,…, Ai+k-1.

Note:

 If K > N, return empty array.
 A[i] is a signed integer

For example,

A=[1, 2, 1, 3, 4, 3] and K = 3

All windows of size K are

[1, 2, 1]
[2, 1, 3]
[1, 3, 4]
[3, 4, 3]

So, we return an array [2, 3, 3, 2].

Solution:

Time: O(n)
Space: O(k)

public class Solution {
    public ArrayList<Integer> dNums(ArrayList<Integer> A, int B) {
        Map<Integer, Integer> map = new HashMap<>();
        ArrayList<Integer> result = new ArrayList<>();
        for (int i = 0; i < A.size(); i ++) {
            int val = A.get(i);
            int freq = map.getOrDefault(val, 0);
            map.put(val, freq + 1);
            if (i >= B) {
                int oldVal = A.get(i - B);
                map.put(oldVal, map.get(oldVal) - 1);
                if (map.get(oldVal) == 0) {
                    map.remove(oldVal);
                }
            }
            // System.out.println(map);
            if (i >= B - 1) result.add(map.size());
        }
        return result;
    }
}