Distinct Numbers in Window

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

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

NOTE: if B > N, return an empty array.

Input Format

First argument is an integer array A Second argument is an integer B. Output Format

Return an integer array. Example Input

Input 1:

 A = [1, 2, 1, 3, 4, 3]
 B = 3

Input 2:

 A = [1, 1, 2, 2]
 B = 1

Example Output

Output 1:

 [2, 3, 3, 2]

Output 2:

 [1, 1, 1, 1]

Example Explanation

Explanation 1:

 A=[1, 2, 1, 3, 4, 3] and B = 3
 All windows of size B are
 [1, 2, 1]
 [2, 1, 3]
 [1, 3, 4]
 [3, 4, 3]
 So, we return an array [2, 3, 3, 2].

Explanation 2:

 Window size is 1, so the output array is [1, 1, 1, 1].

public class Solution {
    public ArrayList<Integer> dNums(ArrayList<Integer> A, int B) {
        if (B > A.size())
            return new ArrayList<Integer>();
        HashMap<Integer, Integer> map = new HashMap<>();
        int unique = 0, i = 0;
        ArrayList<Integer> ans = new ArrayList<>();
        for (i = 0; i < A.size(); i++) {
            if (!map.containsKey(A.get(i))) {
                map.put(A.get(i), 1);
                unique++;
            } else {
                if (map.get(A.get(i)) == 0)
                    unique++;
                map.put(A.get(i), map.get(A.get(i)) + 1);
            }
            if (i >= B - 1) {
                if ((i + 1) - B > 0) {
                    map.put(A.get(i - B), map.get(A.get(i - B)) - 1);
                    if (map.get(A.get(i - B)) == 0)
                        unique--;
                }
                ans.add(unique);
            }
        }
        return ans;
    }
}