First negative integer in every window of size k.cpp

/*
First negative integer in every window of size k
==================================================

Given an array A[] of size N and a positive integer K, find the first negative integer for each and every window(contiguous subarray) of size K.

Example 1:
Input : 
N = 5
A[] = {-8, 2, 3, -6, 10}
K = 2
Output : 
-8 0 -6 -6
Explanation :
First negative integer for each window of size k
{-8, 2} = -8
{2, 3} = 0 (does not contain a negative integer)
{3, -6} = -6
{-6, 10} = -6

Example 2:
Input : 
N = 8
A[] = {12, -1, -7, 8, -15, 30, 16, 28}
K = 3
Output :
-1 -1 -7 -15 -15 0 

Your Task:  
You don't need to read input or print anything. Your task is to complete the function printFirstNegativeInteger() which takes the array A[], its size N and an integer K as inputs and returns the first negative number in every window of size K starting from the first till the end. If a window does not contain a negative integer , then return 0 for that window.


Expected Time Complexity: O(N)
Expected Auxiliary Space: O(K)

Constraints:
1 <= N <= 105
1 <= A[i] <= 105
1 <= K <= N
*/

#define ll long long
vector<ll> printFirstNegativeInteger(ll arr[], ll n, ll k)
{
  vector<ll> ans;
  queue<int> idx;

  for (int i = 0; i < k; ++i)
  {
    if (arr[i] < 0)
      idx.push(i);
  }

  if (idx.size())
    ans.push_back(arr[idx.front()]);
  else
    ans.push_back(0);

  for (int i = k; i < n; ++i)
  {
    while (idx.size() && idx.front() <= i - k)
      idx.pop();
    if (arr[i] < 0)
      idx.push(i);
    if (idx.size())
      ans.push_back(arr[idx.front()]);
    else
      ans.push_back(0);
  }

  return ans;
}