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花花酱 LeetCode 1825. Finding MK Average

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:

  1. 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.
  2. Remove the smallest k elements and the largest k elements from the container.
  3. 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 <= 105
  • 1 <= k*2 < m
  • 1 <= num <= 105
  • At most 105 calls will be made to addElement and calculateMKAverage.

Solution 1: Multiset * 3

Use three multiset to track the left part (smallest k elements), right part (largest k elements) and mid (middle part of m – 2*k elements).

Time complexity: addElememt: O(logn), average: O(1)
Space complexity: O(n)

C++

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