295. 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

Since the data in the data stream is not in order, we should first think of a way to make it in ordered. If we use a vector to save the data stream, we must sort the array every time a new data comes in, which is not efficient.

Use large and small heaps to solve the problem, where the large heap holds the larger numbers in the right half, and the small heap holds the smaller numbers in the left half. In this way, the entire array is divided into two sections in the middle. Since the heap is saved from large to small, we hope that the data in the large pile is from small to large, so it is convenient to take the first one to calculate the median.

Follow up

Q-1: If all integer numbers from the stream are between 0 and 100, how would you optimize it?

We can maintain an integer array of length 100 to store the count of each number along with a total count. Then, we can iterate over the array to find the middle value to get our median. Time and space complexity would be O(100) = O(1).

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

In this case, we can keep a counter for lessThanHundred and greaterThanHundred. As we know the solution will be definitely in 0-100 we don’t need to know those number which are >100 or <0, only count of them will be enough.

class MedianFinder {
    private PriorityQueue<Integer> q1 = new PriorityQueue<>();
    private PriorityQueue<Integer> q2 = new PriorityQueue<>(Collections.reverseOrder());

    /** initialize your data structure here. */
    public MedianFinder() {
    }

    public void addNum(int num) {
        q1.offer(num);
        q2.offer(q1.poll());
        if (q2.size() - q1.size() > 1) {
            q1.offer(q2.poll());
        }
    }

    public double findMedian() {
        if (q2.size() > q1.size()) {
            return q2.peek();
        }
        return (q1.peek() + q2.peek()) * 1.0 / 2;
    }
}

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

class MedianFinder {
public:
    /** initialize your data structure here. */
    MedianFinder() {
    }

    void addNum(int num) {
        q1.push(num);
        q2.push(q1.top());
        q1.pop();
        if (q2.size() - q1.size() > 1) {
            q1.push(q2.top());
            q2.pop();
        }
    }

    double findMedian() {
        if (q2.size() > q1.size()) {
            return q2.top();
        }
        return (double) (q1.top() + q2.top()) / 2;
    }

private:
    priority_queue<int, vector<int>, greater<int>> q1;
    priority_queue<int> q2;
};

/**
 * Your MedianFinder object will be instantiated and called as such:
 * MedianFinder* obj = new MedianFinder();
 * obj->addNum(num);
 * double param_2 = obj->findMedian();
 */

from heapq import heappush, heappop

class MedianFinder:

	def __init__(self):
        # the smaller half of the list, max heap (invert min-heap)
		self.smallerHalf = []
        # the larger half of the list, min heap
		self.largerHalf = []

	def addNum(self, num: int) -> None:

		# trick for smaller half, use -1*val
		# heapq in python does NOT have comparator like in Java
		heappush(self.smallerHalf, -num)
		heappush(self.largerHalf, -heappop(self.smallerHalf)) # note: not self.smallerHalf.pop()

		if len(self.smallerHalf) < len(self.largerHalf):
			heappush(self.smallerHalf, -heappop(self.largerHalf))

	def findMedian(self) -> float:
		if len(self.smallerHalf) == len(self.largerHalf):
			return (-self.smallerHalf[0] + self.largerHalf[0]) / 2.0
		else:
			return float(-self.smallerHalf[0])


# Your MedianFinder object will be instantiated and called as such:
# obj = MedianFinder()
# obj.addNum(num)
# param_2 = obj.findMedian()


# follow up
class MedianFinder:
    def __init__(self):
        self.counts = [0] * 101
        self.total = 0

    def addNum(self, num: int) -> None:
        self.counts[num] += 1
        self.total += 1

    def findMedian(self) -> float:
        if self.total % 2 == 0:
            # even number of elements
            middle1 = self.total // 2
            middle2 = middle1 + 1
            count = 0
            i1 = i2 = 0
            for i in range(101):
                count += self.counts[i]
                if count >= middle1 and i1 == 0:
                    i1 = i
                if count >= middle2:
                    i2 = i
                    break
            return (i1 + i2) / 2
        else:
            # odd number of elements
            middle = self.total // 2 + 1
            count = 0
            for i in range(101):
                count += self.counts[i]
                if count >= middle:
                    return i


############

class MedianFinder:
    def __init__(self):
        """
        initialize your data structure here.
        """
        self.h1 = []
        self.h2 = []

    def addNum(self, num: int) -> None:
        heappush(self.h1, num)
        heappush(self.h2, -heappop(self.h1))
        if len(self.h2) - len(self.h1) > 1:
            heappush(self.h1, -heappop(self.h2))

    def findMedian(self) -> float:
        if len(self.h2) > len(self.h1):
            return -self.h2[0]
        return (self.h1[0] - self.h2[0]) / 2

# Your MedianFinder object will be instantiated and called as such:
# obj = MedianFinder()
# obj.addNum(num)
# param_2 = obj.findMedian()

type MedianFinder struct {
	q1 hp
	q2 hp
}

/** initialize your data structure here. */
func Constructor() MedianFinder {
	return MedianFinder{hp{}, hp{} }
}

func (this *MedianFinder) AddNum(num int) {
	heap.Push(&this.q1, num)
	heap.Push(&this.q2, -heap.Pop(&this.q1).(int))
	if this.q2.Len()-this.q1.Len() > 1 {
		heap.Push(&this.q1, -heap.Pop(&this.q2).(int))
	}
}

func (this *MedianFinder) FindMedian() float64 {
	if this.q2.Len() > this.q1.Len() {
		return -float64(this.q2.IntSlice[0])
	}
	return float64(this.q1.IntSlice[0]-this.q2.IntSlice[0]) / 2.0
}

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

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
func (h *hp) Push(v any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}

class MedianFinder {
    private nums: number[];

    constructor() {
        this.nums = [];
    }

    addNum(num: number): void {
        const { nums } = this;
        let l = 0;
        let r = nums.length;
        while (l < r) {
            const mid = (l + r) >>> 1;
            if (nums[mid] < num) {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
        nums.splice(l, 0, num);
    }

    findMedian(): number {
        const { nums } = this;
        const n = nums.length;
        if ((n & 1) === 1) {
            return nums[n >> 1];
        }
        return (nums[n >> 1] + nums[(n >> 1) - 1]) / 2;
    }
}

/**
 * Your MedianFinder object will be instantiated and called as such:
 * var obj = new MedianFinder()
 * obj.addNum(num)
 * var param_2 = obj.findMedian()
 */

/**
 * initialize your data structure here.
 */
var MedianFinder = function () {
    this.val = [];
};

/**
 * @param {number} num
 * @return {void}
 */
MedianFinder.prototype.addNum = function (num) {
    let left = 0;
    let right = this.val.length;
    while (left < right) {
        let mid = left + ~~((right - left) / 2);
        if (num > this.val[mid]) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    this.val.splice(left, 0, num);
};

/**
 * @return {number}
 */
MedianFinder.prototype.findMedian = function () {
    let mid = ~~(this.val.length / 2);
    return this.val.length % 2 ? this.val[mid] : (this.val[mid - 1] + this.val[mid]) / 2;
};

public class MedianFinder {
    private List<int> nums;
    private int curIndex;

    /** initialize your data structure here. */
    public MedianFinder() {
        nums = new List<int>();
    }

    private int FindIndex(int val) {
        int left = 0;
        int right = nums.Count - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (val > nums[mid]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }
    
    public void AddNum(int num) {
        if (nums.Count == 0) {
            nums.Add(num);
            curIndex = 0;
        } else {
            curIndex = FindIndex(num);
            if (curIndex == nums.Count) {
                nums.Add(num);
            } else {
                nums.Insert(curIndex, num);
            }
        }
    }
    
    public double FindMedian() {
        if (nums.Count % 2 == 1) {
            return (double)nums[nums.Count / 2];
        } else {
            if (nums.Count == 0) {
                return 0;
            } else {
                return (double) (nums[nums.Count / 2 - 1] + nums[nums.Count / 2]) / 2;
            }
        }
    }
}

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

struct MedianFinder {
    nums: Vec<i32>,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl MedianFinder {
    /** initialize your data structure here. */
    fn new() -> Self {
        Self { nums: Vec::new() }
    }

    fn add_num(&mut self, num: i32) {
        let mut l = 0;
        let mut r = self.nums.len();
        while l < r {
            let mid = (l + r) >> 1;
            if self.nums[mid] < num {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
        self.nums.insert(l, num);
    }

    fn find_median(&self) -> f64 {
        let n = self.nums.len();
        if (n & 1) == 1 {
            return f64::from(self.nums[n >> 1]);
        }
        f64::from(self.nums[n >> 1] + self.nums[(n >> 1) - 1]) / 2.0
    }
}/**
 * Your MedianFinder object will be instantiated and called as such:
 * let obj = MedianFinder::new();
 * obj.add_num(num);
 * let ret_2: f64 = obj.find_median();
 */

295 - Find Median from Data Stream

295. Find Median from Data Stream

Description

Solutions

Follow up

Q-1: If all integer numbers from the stream are between 0 and 100, how would you optimize it?

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

All Problems

All Solutions

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295. Find Median from Data Stream

Description

Solutions

Follow up

Q-1: If all integer numbers from the stream are between 0 and 100, how would you optimize it?

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

All Problems

All Solutions