时间复杂度 O(nlogn) 是多少？ 我错过了什么吗？

Question

I have a function for the following problem statement:

Remove k least frequent elements to make the remaining ones as least unique ints set.

How is the Time Complexity O(nlogn)? I am calculating O(n + nlogn).

public class KRemoval {
    public static int solution(int k, ArrayList<Integer> arr, int num) {
        Map<Integer, Integer> map = new HashMap();
        for (int a : arr)
            map.put(a, map.getOrDefault(a, 0) + 1);

        PriorityQueue<Integer> q = new PriorityQueue((a, b) -> map.get(a) - map.get(b));
        for (int key : map.keySet())
            q.offer(key);

        while (k > 0 && !q.isEmpty())
            k -= map.get(q.poll());

        return k < 0 ? q.size() + 1 : q.size();
    }
}

Answer 1

n + n*log(n) can be seen as the sum of two distinct functions, that are f(n) = n and g(n) = n*log(n) , respectively. The first function is slower than the second, since the limit lim_{n -> +inf}{f(n) / g(n)} tends to zero.

Therefore, asymptotically (ie when you "move" towards the infinite), the first function does not really influence the behavior of the total function, which will strictly depend on the behavior of n*log(n) . Thus, we say the cost is O(n*log(n)) .