295. Find Median from Data Stream

Problem Description

The MedianFinder class is designed to find the median of a data stream. 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:
• 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?

Python Solution

class MedianFinder:
    def __init__(self):
        # max heap for the first half
        self.small = []
        # min heap for the second half
        self.large = []

    def addNum(self, num: int) -> None:
        # Always add to small (max heap) first
        heapq.heappush(self.small, -num)
        
        # Make sure every num in small <= every num in large
        if self.small and self.large and -self.small[0] > self.large[0]:
            val = -heapq.heappop(self.small)
            heapq.heappush(self.large, val)
            
        # Handle uneven size
        if len(self.small) > len(self.large) + 1:
            val = -heapq.heappop(self.small)
            heapq.heappush(self.large, val)
        if len(self.large) > len(self.small):
            val = heapq.heappop(self.large)
            heapq.heappush(self.small, -val)

    def findMedian(self) -> float:
        if len(self.small) > len(self.large):
            return -self.small[0]
        return (-self.small[0] + self.large[0]) / 2

Java Solution

class MedianFinder {
    private PriorityQueue small;  // max heap
    private PriorityQueue large;  // min heap
    
    public MedianFinder() {
        small = new PriorityQueue<>((a, b) -> b - a);  // max heap
        large = new PriorityQueue<>();  // min heap
    }
    
    public void addNum(int num) {
        // Always add to small (max heap) first
        small.offer(num);
        
        // Make sure every num in small <= every num in large
        if (!small.isEmpty() && !large.isEmpty() && small.peek() > large.peek()) {
            large.offer(small.poll());
        }
        
        // Handle uneven size
        if (small.size() > large.size() + 1) {
            large.offer(small.poll());
        }
        if (large.size() > small.size()) {
            small.offer(large.poll());
        }
    }
    
    public double findMedian() {
        if (small.size() > large.size()) {
            return small.peek();
        }
        return (small.peek() + large.peek()) / 2.0;
    }
}

C++ Solution

class MedianFinder {
private:
    priority_queue small;  // max heap
    priority_queue, greater> large;  // min heap
    
public:
    MedianFinder() {}
    
    void addNum(int num) {
        // Always add to small (max heap) first
        small.push(num);
        
        // Make sure every num in small <= every num in large
        if (!small.empty() && !large.empty() && small.top() > large.top()) {
            large.push(small.top());
            small.pop();
        }
        
        // Handle uneven size
        if (small.size() > large.size() + 1) {
            large.push(small.top());
            small.pop();
        }
        if (large.size() > small.size()) {
            small.push(large.top());
            large.pop();
        }
    }
    
    double findMedian() {
        if (small.size() > large.size()) {
            return small.top();
        }
        return (small.top() + large.top()) / 2.0;
    }
};

JavaScript Solution

var MedianFinder = function() {
    this.nums = [];
};

/** 
 * @param {number} num
 * @return {void}
 */
MedianFinder.prototype.addNum = function(num) {
    if (this.nums.length === 0) {
        this.nums.push(num);
        return;
    }
    
    // Binary search to find insertion position
    let left = 0;
    let right = this.nums.length;
    while (left < right) {
        let mid = Math.floor((left + right) / 2);
        if (this.nums[mid] < num) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    
    this.nums.splice(left, 0, num);
};

/**
 * @return {number}
 */
MedianFinder.prototype.findMedian = function() {
    const n = this.nums.length;
    if (n % 2 === 1) {
        return this.nums[Math.floor(n/2)];
    }
    return (this.nums[n/2-1] + this.nums[n/2]) / 2;
};

C# Solution

public class MedianFinder {
    private PriorityQueue small;  // max heap
    private PriorityQueue large;  // min heap
    
    public MedianFinder() {
        small = new PriorityQueue();  // max heap
        large = new PriorityQueue();  // min heap
    }
    
    public void AddNum(int num) {
        // Always add to small (max heap) first
        small.Enqueue(num, -num);
        
        // Make sure every num in small <= every num in large
        if (small.Count > 0 && large.Count > 0 && -small.Peek() > large.Peek()) {
            large.Enqueue(-small.Dequeue(), -small.Dequeue());
        }
        
        // Handle uneven size
        if (small.Count > large.Count + 1) {
            large.Enqueue(-small.Dequeue(), -small.Dequeue());
        }
        if (large.Count > small.Count) {
            small.Enqueue(large.Dequeue(), -large.Dequeue());
        }
    }
    
    public double FindMedian() {
        if (small.Count > large.Count) {
            return -small.Peek();
        }
        return (-small.Peek() + large.Peek()) / 2.0;
    }
}