LeetCode 0295. Find Median from Data Stream Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

CoderIndeed
0
0295. Find Median from Data Stream

Description

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

  • For example, for arr = [2,3,4], the median is 3.
  • For example, for arr = [2,3], the median is (2 + 3) / 2 = 2.5.

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. Answers within 10-5 of the actual answer will be accepted.

 

Example 1:

Input
["MedianFinder", "addNum", "addNum", "findMedian", "addNum", "findMedian"]
[[], [1], [2], [], [3], []]
Output
[null, null, null, 1.5, null, 2.0]

Explanation
MedianFinder medianFinder = new MedianFinder();
medianFinder.addNum(1);    // arr = [1]
medianFinder.addNum(2);    // arr = [1, 2]
medianFinder.findMedian(); // return 1.5 (i.e., (1 + 2) / 2)
medianFinder.addNum(3);    // arr[1, 2, 3]
medianFinder.findMedian(); // return 2.0

 

Constraints:

  • -105 <= num <= 105
  • There will be at least one element in the data structure before calling findMedian.
  • At most 5 * 104 calls will be made to addNum and findMedian.

 

Follow up:

  • If all integer numbers from the stream are in the range [0, 100], how would you optimize your solution?
  • If 99% of all integer numbers from the stream are in the range [0, 100], how would you optimize your solution?

Solutions

Solution 1: Min Heap and Max Heap (Priority Queue)

We can use two heaps to maintain all the elements, a min heap minQ and a max heap maxQ, where the min heap minQ stores the larger half, and the max heap maxQ stores the smaller half.

When calling the addNum method, we first add the element to the max heap maxQ, then pop the top element of maxQ and add it to the min heap minQ. If at this time the size difference between minQ and maxQ is greater than 1, we pop the top element of minQ and add it to maxQ. The time complexity is O(log n).

When calling the findMedian method, if the size of minQ is equal to the size of maxQ, it means the total number of elements is even, and we can return the average value of the top elements of minQ and maxQ; otherwise, we return the top element of minQ. The time complexity is O(1).

The space complexity is O(n), where n is the number of elements.

PythonJavaC++GoTypeScriptRustJavaScriptC#Swift
class MedianFinder: def __init__(self): self.minq = [] self.maxq = [] def addNum(self, num: int) -> None: heappush(self.minq, -heappushpop(self.maxq, -num)) if len(self.minq) - len(self.maxq) > 1: heappush(self.maxq, -heappop(self.minq)) def findMedian(self) -> float: if len(self.minq) == len(self.maxq): return (self.minq[0] - self.maxq[0]) / 2 return self.minq[0] # Your MedianFinder object will be instantiated and called as such: # obj = MedianFinder() # obj.addNum(num) # param_2 = obj.findMedian()(code-box)

Post a Comment

0Comments

Post a Comment (0)

#buttons=(Accept !) #days=(20)

Our website uses cookies to enhance your experience. Check Now
Accept !