To find out if a number is present as a result or not, you need to know each other's list if the numbers are above, as well as the numbers below. A list where there are both numbers above and below is not a problem, since you can choose a number in them at your discretion. The problem is with lists where there are only numbers above or only numbers below. The first should be no more than NK, and the second no more than K. If this is not the case, your number cannot be selected. If so, you can always select numbers from lists where there are both numbers above and below so that your number is selected.
This can be checked in linear time, or even better if your first to sort your lists, which gives the overall complexity of O (n.log (n)), where n is the number of numbers in total. Without sorting, you got complexity n².
In your example with lists:
{2 5 3}, {8 1 6}, {7 4 9}
They say that we are looking for the 2-greatest number. For each number, we ask if it can be shouted by the administrator. For each of them, we look to see if the other list has both numbers below and numbers above. Let's look further at some of them.
For 5: in both lists there are numbers above and below. So "5" can be called by the administrator.
For "2": in the second list there are numbers above and below, so I can freely choose the number indicated above or below in this list. But in the third list there are only numbers, so the selected number in this list will always be greater. Since I can freely choose the number lower in the second list, thereby making my “2” the second highest number.
For "1": the second list contains only numbers, so "1" will always be the smallest element.
For "9": it's the other way around, it's always the biggest.