### Problem Statement :

```The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then:

If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set  { 1, 2, 3 } , 2  is the median.
If your set contains an even number of elements, the median is the average of the two middle elements of the sorted sample. In the sorted set { 1, 2, 3 , 4}, 2 + 3 /2,  2.5 is the median.

Given an input stream of n integers, you must perform the following task for each ith integer:

Add the ith integer to a running list of integers.
Find the median of the updated list (i.e., for the first element through the ith  element).
Print the list's updated median on a new line. The printed value must be a double-precision number scaled to 1 decimal place (i.e., 12.3 format).

Input Format

The first line contains a single integer, n, denoting the number of integers in the data stream.
Each line i of the n subsequent lines contains an integer, ai, to be added to your list.

Constraints

1 < = n  <= 10 ^5
0 < = ai  <= 10 ^5

Output Format

After each new integer is added to the list, print the list's updated median on a new line as a single double-precision number scaled to 1 decimal place (i.e., 12.3 format).```

### Solution :

```                            ```Solution in C :

In C ++ :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std;

int main() {

size_t n;
cin >> n;

vector<size_t> numbers;
for (size_t i = 0; i < n; ++i){
size_t number;
cin >> number;

// keep numbers sorted
numbers.insert(lower_bound(numbers.begin(), numbers.end(), number), number);

double median;
if (i % 2 == 0){ // odd number of elements
median = numbers[i / 2];
}
else{ // even
median = (double(numbers[i / 2]) + numbers[i / 2 + 1]) / 2;
}
cout << fixed << setprecision(1) << median << "\n";
}
return 0;
}

In Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static class MinComparator implements Comparator<Integer> {
@Override
public int compare(Integer i1, Integer i2) {
if (i1 < i2) {
return -1;
} else if (i1 > i2) {
return 1;
} else {
return 0;
}
}
}

public static class MaxComparator implements Comparator<Integer> {
@Override
public int compare(Integer i1, Integer i2) {
if (i1 > i2) {
return -1;
} else if (i1 < i2) {
return 1;
} else {
return 0;
}
}
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);

int n = in.nextInt();

Queue<Integer> minHeap = new PriorityQueue<>(n/2+1, new MinComparator());
Queue<Integer> maxHeap = new PriorityQueue<>(n/2+1, new MaxComparator());

for (int i = 0; i < n; i++) {
int num = in.nextInt();

if (!maxHeap.isEmpty() && num > maxHeap.peek()) {
minHeap.offer(num);
} else {
maxHeap.offer(num);
}

if (minHeap.size() > maxHeap.size() + 1) {
maxHeap.offer(minHeap.poll());
} else if (maxHeap.size() > minHeap.size() + 1) {
minHeap.offer(maxHeap.poll());
}

double median = 0.0;
if (minHeap.size() == maxHeap.size()) {
median = 0.5 * (minHeap.peek() + maxHeap.peek());
} else {
median = (minHeap.size() > maxHeap.size()) ? minHeap.peek() : maxHeap.peek();
}

System.out.println(String.format("%1.1f", median));
}
}

}

In C :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

int a, n, i, j, t, k, start, end, mid;
scanf("%d", &n);
for(i=0; i < n; i++) {
scanf("%d", &t);
start = 0;
end = i;
mid = (end-start)/2;
while (start != mid && mid != end) {

if (a[mid] > t)
end = mid;
else if (a[mid] <= t)
start = mid;
mid = start + (end - start)/2;

}

for(j=mid; j < i; j++) {
if (a[j] > t) {
break;
}
}

//memcpy(&a[j+1], &a[j], (i-j)*sizeof(int));
for(k = i; k >= (j+1); k--)
a[k] = a[k-1];
a[j] = t;
//for(j = 0; j <= i; j++)
//    printf("%d ", a[j]);
//printf("\n");
if(i%2 == 0) {
printf("%.1f\n", (float)a[i/2]);
} else {
printf("%.1f\n", (float)(a[i/2] + a[(i+1)/2])/2.0);
}
}
return 0;
}

In Python3 :

import heapq as hq

minheap = []
maxheap = []
N = int(input())
for i in range(0, N):
x = int(input())
if i % 2 == 0:
hq.heappush(maxheap, -1*x)
if len(minheap) == 0:
x = -1*maxheap
print(float(x))
continue
elif -1*maxheap > minheap:
toMin = -1*hq.heappop(maxheap)
toMax = hq.heappop(minheap)
hq.heappush(minheap, toMin)
hq.heappush(maxheap, -1*toMax)
x = -1*maxheap
print(float(x))
else:
toMin = -1*hq.heappushpop(maxheap, -1*x)
hq.heappush(minheap, toMin)
x = (-1*maxheap+minheap)/2.0
print(x)```
```

## Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

## Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

## Sparse Arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun

## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

## Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

## Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink