1825. Finding MK Average

Problem

You are given two integers, m and k, and a stream of integers. You are tasked to implement a data structure that calculates the MKAverage for the stream.

The MKAverage can be calculated using these steps:

• If the number of the elements in the stream is less than m you should consider the MKAverage to be -1. Otherwise, copy the last m elements of the stream to a separate container.

• Remove the smallest k elements and the largest k elements from the container.

• Calculate the average value for the rest of the elements rounded down to the nearest integer.

Implement the MKAverage class:

• MKAverage(int m, int k) Initializes the MKAverage object with an empty stream and the two integers m and k.

• void addElement(int num) Inserts a new element num into the stream.

• int calculateMKAverage() Calculates and returns the MKAverage for the current stream rounded down to the nearest integer.

  Example 1:

Input
["MKAverage", "addElement", "addElement", "calculateMKAverage", "addElement", "calculateMKAverage", "addElement", "addElement", "addElement", "calculateMKAverage"]
[[3, 1], [3], [1], [], [10], [], [5], [5], [5], []]
Output
[null, null, null, -1, null, 3, null, null, null, 5]

Explanation
MKAverage obj = new MKAverage(3, 1); 
obj.addElement(3);        // current elements are [3]
obj.addElement(1);        // current elements are [3,1]
obj.calculateMKAverage(); // return -1, because m = 3 and only 2 elements exist.
obj.addElement(10);       // current elements are [3,1,10]
obj.calculateMKAverage(); // The last 3 elements are [3,1,10].
                          // After removing smallest and largest 1 element the container will be [3].
                          // The average of [3] equals 3/1 = 3, return 3
obj.addElement(5);        // current elements are [3,1,10,5]
obj.addElement(5);        // current elements are [3,1,10,5,5]
obj.addElement(5);        // current elements are [3,1,10,5,5,5]
obj.calculateMKAverage(); // The last 3 elements are [5,5,5].
                          // After removing smallest and largest 1 element the container will be [5].
                          // The average of [5] equals 5/1 = 5, return 5

  Constraints:

• 3 <= m <= 10^5

• 1 <= k*2 < m

• 1 <= num <= 10^5

• At most 10^5 calls will be made to addElement and calculateMKAverage.

Solution

class MKAverage {
    private final double m;
    private final double k;
    private final double c;
    private double avg;
    private final Bst middle;
    private final Bst min;
    private final Bst max;
    private final Deque<Integer> q;

    public MKAverage(int m, int k) {
        this.m = m;
        this.k = k;
        this.c = m - k * 2;
        this.avg = 0;
        this.middle = new Bst();
        this.min = new Bst();
        this.max = new Bst();
        this.q = new ArrayDeque<>();
    }

    public void addElement(int num) {
        if (min.size < k) {
            min.add(num);
            q.offer(num);
            return;
        }
        if (max.size < k) {
            min.add(num);
            max.add(min.removeMax());
            q.offer(num);
            return;
        }

        if (num >= min.lastKey() && num <= max.firstKey()) {
            middle.add(num);
            avg += num / c;
        } else if (num < min.lastKey()) {
            min.add(num);
            int val = min.removeMax();
            middle.add(val);
            avg += val / c;
        } else if (num > max.firstKey()) {
            max.add(num);
            int val = max.removeMin();
            middle.add(val);
            avg += val / c;
        }

        q.offer(num);

        if (q.size() > m) {
            num = q.poll();
            if (middle.containsKey(num)) {
                avg -= num / c;
                middle.remove(num);
            } else if (min.containsKey(num)) {
                min.remove(num);
                int val = middle.removeMin();
                avg -= val / c;
                min.add(val);
            } else if (max.containsKey(num)) {
                max.remove(num);
                int val = middle.removeMax();
                avg -= val / c;
                max.add(val);
            }
        }
    }

    public int calculateMKAverage() {
        if (q.size() < m) {
            return -1;
        }
        return (int) avg;
    }

    static class Bst {
        TreeMap<Integer, Integer> map;
        int size;

        public Bst() {
            this.map = new TreeMap<>();
            this.size = 0;
        }

        void add(int num) {
            int count = map.getOrDefault(num, 0) + 1;
            map.put(num, count);
            size++;
        }

        void remove(int num) {
            int count = map.getOrDefault(num, 1) - 1;
            if (count > 0) {
                map.put(num, count);
            } else {
                map.remove(num);
            }
            size--;
        }

        int removeMin() {
            int key = map.firstKey();

            remove(key);

            return key;
        }

        int removeMax() {
            int key = map.lastKey();

            remove(key);

            return key;
        }

        boolean containsKey(int key) {
            return map.containsKey(key);
        }

        int firstKey() {
            return map.firstKey();
        }

        int lastKey() {
            return map.lastKey();
        }
    }
}

/**
 * Your MKAverage object will be instantiated and called as such:
 * MKAverage obj = new MKAverage(m, k);
 * obj.addElement(num);
 * int param_2 = obj.calculateMKAverage();
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).