Rolling Median - Amazon Top Interview Questions

Problem Statement :

```Implement a RollingMedian class with the following methods:

median() which retrieves the current median of all numbers added
Median of [1, 2, 3] is 2 whereas median of [1, 2, 3, 4] is 2.5.

Constraints

n ≤ 100,000 where n is the number of calls to add and median

Example 1

Input

arguments = [[], [1], [2], [3], [], [4], []]`

Output

[None, None, None, None, 2, None, 2.5]

Explanation

We first add 1, 2, and 3. The median is then 2. Then we add 4. Median is now (2 + 3) / 2 = 2.5```

Solution :

```                        ```Solution in C++ :

class RollingMedian {
public:
multiset<int> order;
multiset<int>::iterator it;
RollingMedian() {
}

order.insert(val);
if (order.size() == 1) {
it = order.begin();
} else {
if (val < *it and order.size() % 2 == 0) {
--it;
}
if (val >= *it and order.size() % 2 != 0) {
++it;
}
}
}

double median() {
if (order.size() % 2 != 0) {
return double(*it);
} else {
auto one = *it, two = *next(it);
return double(one + two) / 2.0;
}
}
};```
```

```                        ```Solution in Java :

import java.util.*;

class RollingMedian {
PriorityQueue<Integer> small, great;

public RollingMedian() {
this.small = new PriorityQueue<Integer>(Collections.reverseOrder());
this.great = new PriorityQueue<Integer>();
}

if (this.small.size() == 0 && this.great.size() == 0) {
} else if (this.small.size() > this.great.size()) {
if (this.small.peek() > val) {
} else {
}

} else {
if (this.small.peek() >= val) {
} else {
}
}
}

public double median() {
if (this.small.size() > this.great.size()) {
return this.small.peek();
} else {
return (double) (this.small.peek() + this.great.peek()) / 2;
}
}
}```
```

```                        ```Solution in Python :

class RollingMedian:
def __init__(self):
self.l = SortedList()

def median(self):
if len(self.l) % 2 == 0:

return (self.l[len(self.l) // 2] + self.l[(len(self.l) // 2) - 1]) / 2
else:
return self.l[len(self.l) // 2]```
```

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