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Java Solution for LeetCode algorithm problems, continually updating
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1825\. Finding MK Average
Hard
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`.