295. Find Median from Data Stream

Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value.

For example,

[2,3,4], the median is 3

[2,3], the median is (2 + 3) / 2 = 2.5

Design a data structure that supports the following two operations:

• void addNum(int num) - Add a integer number from the data stream to the data structure.
• double findMedian() - Return the median of all elements so far.

Example:

addNum(1)
addNum(2)
findMedian() -> 1.5
addNum(3)
findMedian() -> 2

Follow up:

1. If all integer numbers from the stream are between 0 and 100, how would you optimize it?
2. If 99% of all integer numbers from the stream are between 0 and 100, how would you optimize it?

# @lc code=start
using LeetCode

using DataStructures

Base.@kwdef struct MedianFinder
    maxheap::BinaryMaxHeap{Int} = BinaryMaxHeap{Int}()
    minheap::BinaryMinHeap{Int} = BinaryMinHeap{Int}()
end
# function Base.show(io::IO, heap::MedianFinder)
#     return print(io, "$(heap.maxheap.valtree):max -- min:$(reverse(heap.minheap.valtree))")
# end

function add_num!(heap::MedianFinder, num::Int)
    # maximal element of maxheap <= minimal element of minheap
    # length of maxheap <= length of minheap
    hi, ha = heap.minheap, heap.maxheap
    if length(ha) < length(hi)
        # e.g. [2]--[4,5] =6> [2]--[4,5,6] => [2,4]--[5,6]
        push!(hi, num)
        push!(ha, pop!(hi))
    else
        # e.g. [4]--[5] =2> [2,4]--[5] => [2]--[4,5]
        push!(ha, num)
        push!(hi, pop!(ha))
    end
    return nothing
end

function find_median(heap::MedianFinder)::Float64
    hi, ha = heap.minheap, heap.maxheap
    return length(hi) == length(ha) ? (first(hi) + first(ha)) / 2 : first(hi)
end
# @lc code=end

find_median (generic function with 1 method)

This page was generated using DemoCards.jl and Literate.jl.