Distinct Numbers in Window.cpp

/*
Distinct Numbers in Window
==========================

Problem Description

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].
*/

vector<int> Solution::dNums(vector<int> &A, int B)
{
  if (B > A.size())
    return {};

  unordered_map<int, int> freq;
  unordered_set<int> dist;

  vector<int> ans;
  int i = 0;

  while (i < B)
  {
    freq[A[i]]++;
    dist.insert(A[i]);
    i++;
  }
  ans.push_back(dist.size());

  int j = 0;
  while (i < A.size())
  {
    freq[A[j]]--;
    if (freq[A[j]] == 0)
      dist.erase(A[j]);
    freq[A[i]]++;
    dist.insert(A[i]);
    ans.push_back(dist.size());
    i++;
    j++;
  }

  return ans;
}