LeetCode in Net

295. Find Median from Data Stream

Hard

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.

Implement the MedianFinder class:

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:

Follow up:

Solution

using System;
using System.Collections.Generic;

public class MedianFinder {
    private PriorityQueue<int, int> left = new();
    private PriorityQueue<int, int> right = new();
    private bool odd = false;

    public void AddNum(int n) {
        odd = !odd;
        int m = right.EnqueueDequeue(n, -n);
        left.Enqueue(m, m);     
        if (left.Count - 1 > right.Count) {
            m = left.Dequeue();
            right.Enqueue(m, -m);
        }
    }
    
    public double FindMedian() =>
        odd ? left.Peek() : (left.Peek() + right.Peek()) / 2.0;
}

/**
 * Your MedianFinder object will be instantiated and called as such:
 * MedianFinder obj = new MedianFinder();
 * obj.AddNum(num);
 * double param_2 = obj.FindMedian();
}
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