Count distinct elements in every window

PROBLEM :
Given an array A[] of size n and an integer k, your task is to complete the function countDistinct which prints the count of distinct numbers in all windows of size k in the array A[].

Example

Input:  A[] = {1, 2, 1, 3, 4, 2, 3};
            k = 4
Output:
3 4 4 3
Explanation:
First window is {1, 2, 1, 3}, count of distinct numbers is 3
Second window is {2, 1, 3, 4} count of distinct numbers is 4
Third window is {1, 3, 4, 2} count of distinct numbers is 4
Fourth window is {3, 4, 2, 3} count of distinct numbers is 3


Input:
The first line of input contains an integer T denoting the no of test cases. Then T test cases follow. Each test case contains two integers n and k. Then in the next line are N space separated values of the array A[].

Output:
For each test case in a new line print the space separated values denoting counts of distinct numbers in all windows of size k in the array A[].

Constraints:
1<=T<=100
1<=n,k<=100
1<=A[]<=100

Example(To be used only for expected output):
Input:
2
7 4
1 2 1 3 4 2 3
3 2
4 1 1

Output:
3 4 4 3
2 1


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SIMPLE c++ IMPLEMENTATION :
--------------------------------------------------------------------------------

/*You are required to complete below method */

void countDistinct(int A[], int k, int n)
{
    map<int,int> window ;
    int distinct=0 ;
   
    for(int i=0;i<k;i++)
    {
        if(window[A[i]]==0)
            distinct++ ;
        window[A[i]]+=1 ;
    }
   
    cout<<distinct<<" " ;
   
    for(int i=k;i<n;i++)
    {
        if(window[A[i-k]]==1)
            distinct-- ;
        window[A[i-k]]-=1 ;
       
        if(window[A[i]]==0)
            distinct++ ;
        window[A[i]]+=1 ;
       
        cout<<distinct<<" " ;
    }
}

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