[NeetCode 150] Find Median in a Data Stream

Find Median in a Data Stream

The median is the middle value in a sorted list of integers. For lists of even length, there is no middle value, so the median is the mean of the two middle values.

For example:

For arr = [1,2,3], the median is 2.

For arr = [1,2], the median is (1 + 2) / 2 = 1.5

Implement the MedianFinder class:

MedianFinder() initializes the MedianFinder object.

void addNum(int num) adds the integer num from the data stream to the data structure.

double findMedian() returns the median of all elements so far.

Example 1:

复制代码
Input:
["MedianFinder", "addNum", "1", "findMedian", "addNum", "3" "findMedian", "addNum", "2", "findMedian"]

Output:
[null, null, 1.0, null, 2.0, null, 2.0]

Explanation:

MedianFinder medianFinder = new MedianFinder();

medianFinder.addNum(1); // arr = [1]

medianFinder.findMedian(); // return 1.0

medianFinder.addNum(3); // arr = [1, 3]

medianFinder.findMedian(); // return 2.0

medianFinder.addNum(2); // arr[1, 2, 3]

medianFinder.findMedian(); // return 2.0

Constraints:

复制代码
-100,000 <= num <= 100,000

findMedian will only be called after adding at least one integer to the data structure.

Solution

We can divide this ordered list into 2 parts. Maintain the first half via a max-heap and maintain the second half via a min-heap. If we keep the balance between the size of these two heaps, we can guarantee that the median is always at the top of them.

To do this, we need to adjust the size of heaps after each addNum. As we only add 1 number once, we only need to move at most 1 element from one heap to another.

Code

heapq is a good way to realize heap (or say priority queue).

py 复制代码
class MedianFinder:

    def __init__(self):
        self.first = [(100001, -100001)]
        self.second = [(100001, 100001)]
        

    def addNum(self, num: int) -> None:
        first_max = self.first[0][1]
        second_min = self.second[0][1]
        if num <= first_max:
            heapq.heappush(self.first, (-num, num))
        else:
            heapq.heappush(self.second, (num, num))
        if len(self.first) > len(self.second) + 1:
            temp = heapq.heappop(self.first)
            heapq.heappush(self.second, (-temp[0], temp[1]))
        if len(self.second) > len(self.first):
            temp = heapq.heappop(self.second)
            heapq.heappush(self.first, (-temp[0], temp[1]))
        

    def findMedian(self) -> float:
        if len(self.first) == len(self.second):
            return (self.first[0][1]+self.second[0][1])/2
        else:
            return self.first[0][1]
        
        
相关推荐
HelloDam7 分钟前
基于元素小组的归并排序算法
后端·算法·排序算法
HelloDam7 分钟前
基于连贯性算法的多边形扫描线生成(适用于凸多边形和凹多边形)【原理+java实现】
算法
逸狼32 分钟前
【Java 优选算法】二分算法(下)
数据结构
uhakadotcom1 小时前
Apache Airflow入门指南:数据管道的强大工具
算法·面试·github
跳跳糖炒酸奶2 小时前
第四章、Isaacsim在GUI中构建机器人(2):组装一个简单的机器人
人工智能·python·算法·ubuntu·机器人
绵绵细雨中的乡音2 小时前
动态规划-第六篇
算法·动态规划
程序员黄同学2 小时前
动态规划,如何应用动态规划解决实际问题?
算法·动态规划
march_birds2 小时前
FreeRTOS 与 RT-Thread 事件组对比分析
c语言·单片机·算法·系统架构
斯汤雷3 小时前
Matlab绘图案例,设置图片大小,坐标轴比例为黄金比
数据库·人工智能·算法·matlab·信息可视化
云 无 心 以 出 岫3 小时前
贪心算法QwQ
数据结构·c++·算法·贪心算法