stack to serve as the monotonic stack.additionalCount as the number of elements we can drop, which is len(nums) - k.nums:
a. While stack is not empty and the last element of stack is greater than the current element and additionalCount is greater than 0, pop the last element from stack and decrement additionalCount.
b. If the length of stack is less than k, append the current element to stack.k elements of stack as they form the most competitive subsequence.