g0201_0300.s0295_find_median_from_data_stream.MedianFinder Maven / Gradle / Ivy

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package g0201_0300.s0295_find_median_from_data_stream;

// #Hard #Top_100_Liked_Questions #Top_Interview_Questions #Sorting #Two_Pointers #Design
// #Heap_Priority_Queue #Data_Stream #Big_O_Time_O(n*log_n)_Space_O(n)
// #2022_07_06_Time_151_ms_(80.24%)_Space_125.2_MB_(44.11%)

import java.util.PriorityQueue;

/**
 * 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.
 *
 * *   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?
**/
public class MedianFinder {
    // take two queues one is for storing upper half and the other is for lowerhalf
    // max stores the lower half
    // min heap stores the upper half
    private PriorityQueue maxHeap;
    private PriorityQueue minHeap;

    // initialize your data structure here.
    public MedianFinder() {
        maxHeap = new PriorityQueue<>((a, b) -> (b - a));
        minHeap = new PriorityQueue<>();
    }

    public void addNum(int num) {
        if (maxHeap.isEmpty() || maxHeap.peek() > num) {
            maxHeap.offer(num);
        } else {
            minHeap.offer(num);
        }
        if (Math.abs(maxHeap.size() - minHeap.size()) > 1) {
            balance(maxHeap, minHeap);
        }
    }

    public void balance(PriorityQueue maxHeap, PriorityQueue minHeap) {
        PriorityQueue large = maxHeap.size() > minHeap.size() ? maxHeap : minHeap;
        PriorityQueue small = maxHeap.size() > minHeap.size() ? minHeap : maxHeap;
        small.offer(large.poll());
    }

    public double findMedian() {
        PriorityQueue large = maxHeap.size() > minHeap.size() ? maxHeap : minHeap;
        PriorityQueue small = maxHeap.size() > minHeap.size() ? minHeap : maxHeap;
        if (large.size() == small.size()) {
            return (double) (large.peek() + small.peek()) / 2;
        } else {
            return large.peek();
        }
    }
}

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