Leetcode #1825: Finding MK Average
In this guide, we solve Leetcode #1825 Finding MK Average in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Design, Queue, Data Stream, Ordered Set, Heap (Priority Queue)
Intuition
We need to repeatedly access the smallest or largest element as the input changes.
A heap provides fast insertions and removals while keeping order.
Approach
Push candidates into the heap as you scan, and pop when you need the best element.
Keep the heap size bounded if the problem requires a top-k structure.
Steps:
- Push candidates into a heap.
- Pop the best candidate when needed.
- Maintain heap size or invariants.
Example
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
Python Solution
class MKAverage:
def __init__(self, m: int, k: int):
self.m = m
self.k = k
self.s = 0
self.q = deque()
self.lo = SortedList()
self.mid = SortedList()
self.hi = SortedList()
def addElement(self, num: int) -> None:
if not self.lo or num <= self.lo[-1]:
self.lo.add(num)
elif not self.hi or num >= self.hi[0]:
self.hi.add(num)
else:
self.mid.add(num)
self.s += num
self.q.append(num)
if len(self.q) > self.m:
x = self.q.popleft()
if x in self.lo:
self.lo.remove(x)
elif x in self.hi:
self.hi.remove(x)
else:
self.mid.remove(x)
self.s -= x
while len(self.lo) > self.k:
x = self.lo.pop()
self.mid.add(x)
self.s += x
while len(self.hi) > self.k:
x = self.hi.pop(0)
self.mid.add(x)
self.s += x
while len(self.lo) < self.k and self.mid:
x = self.mid.pop(0)
self.lo.add(x)
self.s -= x
while len(self.hi) < self.k and self.mid:
x = self.mid.pop()
self.hi.add(x)
self.s -= x
def calculateMKAverage(self) -> int:
return -1 if len(self.q) < self.m else self.s // (self.m - 2 * self.k)
# Your MKAverage object will be instantiated and called as such:
# obj = MKAverage(m, k)
# obj.addElement(num)
# param_2 = obj.calculateMKAverage()
Complexity
The time complexity is O(n log n). The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.