- Difficulty: Hard
- Tags: LeetCode, Hard, Design, Queue, Data Stream, Ordered Set, Heap (Priority Queue), leetcode-1825, O(m), Sorted List
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 lastm
elements of the stream to a separate container. - Remove the smallest
k
elements and the largestk
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 integersm
andk
.void addElement(int num)
Inserts a new elementnum
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 toaddElement
andcalculateMKAverage
.