Find Median from 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.
- For example, for
arr = [2,3,4]
, the median is3
. - For example, for
arr = [2,3]
, the median is(2 + 3) / 2 = 2.5
.
Implement the MedianFinder class:
MedianFinder()
initializes theMedianFinder
object.void addNum(int num)
adds the integernum
from the data stream to the data structure.double findMedian()
returns the median of all elements so far. Answers within10-5
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 toaddNum
andfindMedian
.
Find Median from Data Stream Solutions
✅Time: O(n log n )
✅Space: O(n)
C++
class MedianFinder {
public:
void addNum(int num) {
if (maxHeap.empty() || num <= maxHeap.top())
maxHeap.push(num);
else
minHeap.push(num);
// balance two heaps s.t.
// |maxHeap| >= |minHeap| and |maxHeap| - |minHeap| <= 1
if (maxHeap.size() < minHeap.size())
maxHeap.push(minHeap.top()), minHeap.pop();
else if (maxHeap.size() - minHeap.size() > 1)
minHeap.push(maxHeap.top()), maxHeap.pop();
}
double findMedian() {
if (maxHeap.size() == minHeap.size())
return (maxHeap.top() + minHeap.top()) / 2.0;
return maxHeap.top();
}
private:
priority_queue<int> maxHeap;
priority_queue<int, vector<int>, greater<>> minHeap;
};
Java
class MedianFinder {
public void addNum(int num) {
if (maxHeap.isEmpty() || num <= maxHeap.peek())
maxHeap.offer(num);
else
minHeap.offer(num);
// balance two heaps s.t.
// |maxHeap| >= |minHeap| and |maxHeap| - |minHeap| <= 1
if (maxHeap.size() < minHeap.size())
maxHeap.offer(minHeap.poll());
else if (maxHeap.size() - minHeap.size() > 1)
minHeap.offer(maxHeap.poll());
}
public double findMedian() {
if (maxHeap.size() == minHeap.size())
return (double) (maxHeap.peek() + minHeap.peek()) / 2.0;
return (double) maxHeap.peek();
}
private PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
private PriorityQueue<Integer> minHeap = new PriorityQueue<>();
}
Python
class MedianFinder:
def __init__(self):
self.maxHeap = []
self.minHeap = []
def addNum(self, num: int) -> None:
if not self.maxHeap or num <= -self.maxHeap[0]:
heapq.heappush(self.maxHeap, -num)
else:
heapq.heappush(self.minHeap, num)
# balance two heaps s.t.
# |maxHeap| >= |minHeap| and |maxHeap| - |minHeap| <= 1
if len(self.maxHeap) < len(self.minHeap):
heapq.heappush(self.maxHeap, -heapq.heappop(self.minHeap))
elif len(self.maxHeap) - len(self.minHeap) > 1:
heapq.heappush(self.minHeap, -heapq.heappop(self.maxHeap))
def findMedian(self) -> float:
if len(self.maxHeap) == len(self.minHeap):
return (-self.maxHeap[0] + self.minHeap[0]) / 2.0
return -self.maxHeap[0]
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