Approach

We want the ~k~-th smallest after each prefix. A direct sort of every prefix is too slow.

Keep a max-heap containing exactly the ~k~ smallest values seen so far.

• If the heap has fewer than ~k~ values, push the new value.
• Otherwise, compare the new value to the largest inside this set (the heap top).
  • If the new value is smaller, it belongs in the ~k~ smallest set, so replace the top.
  • Otherwise ignore it.

Then:

• if heap size is less than ~k~, answer is -1;
• else the heap top is the current ~k~-th smallest.

In Python, heapq is a min-heap, so we store negatives to simulate a max-heap.

Time complexity is ~O(n \log k)~.

Solution (Python)

import heapq

n, k = map(int, input().split())
values = list(map(int, input().split()))

heap = []
answers = []

for x in values:
    if len(heap) < k:
        heapq.heappush(heap, -x)
    elif x < -heap[0]:
        heapq.heapreplace(heap, -x)

    if len(heap) < k:
        answers.append("-1")
    else:
        answers.append(str(-heap[0]))

print("\n".join(answers))