**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

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`

of the actual answer will be accepted.^{-5}

**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:**

`-10`

^{5}<= num <= 10^{5}- There will be at least one element in the data structure before calling
`findMedian`

. - At most
`5 * 10`

calls will be made to^{4}`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?

```
class MedianFinder {
private Queue<Long> small = new PriorityQueue(),
large = new PriorityQueue();
public void addNum(int num) {
large.add((long) num);
small.add(-large.poll());
if (large.size() < small.size())
large.add(-small.poll());
}
public double findMedian() {
return large.size() > small.size()
? large.peek()
: (large.peek() - small.peek()) / 2.0;
}
};
```

```
class MedianFinder {
priority_queue<long> small, large;
public:
void addNum(int num) {
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;
}
};
```

```
from heapq import *
class MedianFinder:
def __init__(self):
self.heaps = [], []
def addNum(self, num):
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
```

In our experience, we suggest you solve this Find Median from Data Stream LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Find Median from Data Stream LeetCode Solution

I hope this Find Median from Data Stream LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

**More Coding Solutions >>**