Top K most frequent number in a stream

#1

I am not able to understand the problem statement.

  1. It says to print the numbers in increasing order if the frequency is same then do we need to print all the numbers with the same frequency or just the k elements.
  2. Explain the sample output and please tell the output for the similar sample input having the value of k = 3
  3. Do we need to wait till the k elements are taken from the array before giving any output.
#2

@isa67719
In this problem you are supposed to output atmost k integers on every iteration from the input you have taken so far.
Let us dry run the sample input given in the question for better understanding.
I believe this will answer all three of your queries.

Sample Input:
1
5 2
5 1 3 5 2
Sample Output:
5 1 5 1 3 5 1 5 1

Let us say store the input in an array named A

No of testcases = 1
Given n = 5 (We will get 5 integer elements in our array)
k = 2 (We have to output at max 2 integers on every iteration)

Iteration 1 :
Input : 5
A = {5}
Output : 5
Explanation : Since there is only one integer with frequency = 1 , we output it

Iteration 2 :
Input : 1
A = {1,5}
Output : 1 5
Explanation : Since there are two elements in A , we output them in sorted order

Iteration 3 :
Input : 3
A = {1,3,5}
Output : 1 3
Explanation : We output k = 2 elements from the array. Since frequency of all elements is the same , we output them in sorted order and 1 and 3 being the smallest gets printed

Iteration 4 :
Input : 5
A = {1,3,5,5}
Output : 5 1
Explanation : 5 has the highest frequency in the array so it gets printed first . Then the remaining elements having the same frequency get printed in the sorted order. Since we only have to print k=2 elements , only 5 and 1 get printed.

Iteration 5 :
Input : 2
A = {1,2,3,5,5}
Output : 5 1
Explanation : 5 still has the highest frequency and 1 being the smallest of the remaining elements gets to be printed .

Final Output : 5 1 5 1 3 5 1 5 1

I hope this clears your doubt and helps you understand the question.