给定一个大小为n的数字集合，以及一个Q输入列表，每个输入将在该集合中删除或添加一个数字，每次输入后，我都要输出该集合中数字的最小绝对差。
限制条件：
2 <= N <= 10^6
1 <= Q <= 10^6
-10^9 <= set[i] <= 10^9
示例：
输入：

set = [2, 4, 7]
ADD 6
REMOVE 7
REMOVE 4
ADD 2

输出：

1
2
4
0

我的任务是使用时间复杂度为**O（（N+Q）log（N+Q））**或更好的算法来解决这个问题。
我当前的实现不够快，但如下所示：

TreeSet<Integer> tree = new TreeSet<>();
HashMap<Integer, Integer> numberFreq = new HashMap<>();

int dupeCount = 0;

for (int i : set) {
    tree.add(i);
    if (numberFreq.get(i) > 0) dupeCount++;
    numberFreq.put(i, numberFreq.getOrDefault(i, 0) + 1);
}

void add(int i) {
    if (numberFreq.get(i) > 0) dupeCount++;
    numberFreq.put(i, numberFreq.getOrDefault(i, 0) + 1);
    tree.add(i); // if duplicate nothing gets added anyway

    if (dupeCount > 0) Console.write(0);
    else {
        int smallestBall = ballTree.first();
        int absDiff;
        int maxAbsDiff = Integer.MAX_VALUE;
        while (ballTree.higher(smallestBall) != null) {
            absDiff = Math.abs(smallestBall - ballTree.higher(smallestBall));
            maxAbsDiff = Math.min(absDiff, maxAbsDiff);
            smallestBall = ballTree.higher(smallestBall);
        }
    Console.write(maxAbsDiff);
    }

}

void remove(int i) {
     if (numberFreq.get(i) > 0) dupeCount--;
     else tree.remove(i);
     numberFreq.put(i, numberFreq.get(i) - 1);

     if (dupeCount > 0) Console.write(0);
     else {
         int smallestBall = ballTree.first();
         int absDiff;
         int maxAbsDiff = Integer.MAX_VALUE;
         while (ballTree.higher(smallestBall) != null) {
             absDiff = Math.abs(smallestBall - ballTree.higher(smallestBall));
             maxAbsDiff = Math.min(absDiff, maxAbsDiff);
             smallestBall = ballTree.higher(smallestBall);
        }
     Console.write(maxAbsDiff);
     }
}

我已经试了两天了，我完全迷路了。