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# Find Median from Data Stream LeetCode Solution

## Problem – 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 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
[[], [1], [2], [], [3], []]
Output
[null, null, null, 1.5, null, 2.0]

Explanation
MedianFinder medianFinder = new MedianFinder();
medianFinder.addNum(2);    // arr = [1, 2]
medianFinder.findMedian(); // return 1.5 (i.e., (1 + 2) / 2)
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`.

• 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?

### Find Median from Data Stream LeetCode Solution in Java

``````class MedianFinder {

private Queue<Long> small = new PriorityQueue(),
large = new PriorityQueue();

if (large.size() < small.size())
}

public double findMedian() {
return large.size() > small.size()
? large.peek()
: (large.peek() - small.peek()) / 2.0;
}
};
``````

### Find Median from Data Stream LeetCode Solution in C++

``````class MedianFinder {
priority_queue<long> small, large;
public:

small.push(num);
large.push(-small.top());
small.pop();
if (small.size() < large.size()) {
small.push(-large.top());
large.pop();
}
}

double findMedian() {
return small.size() > large.size()
? small.top()
: (small.top() - large.top()) / 2.0;
}
};
``````

### Find Median from Data Stream LeetCode Solution in Python

``````from heapq import *

class MedianFinder:

def __init__(self):
self.heaps = [], []

small, large = self.heaps
heappush(small, -heappushpop(large, num))
if len(large) < len(small):
heappush(large, -heappop(small))

def findMedian(self):
small, large = self.heaps
if len(large) > len(small):
return float(large[0])
return (large[0] - small[0]) / 2.0
``````
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