**Find the Running Median**

### 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[100000], 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[0]
print(float(x))
continue
elif -1*maxheap[0] > minheap[0]:
toMin = -1*hq.heappop(maxheap)
toMax = hq.heappop(minheap)
hq.heappush(minheap, toMin)
hq.heappush(maxheap, -1*toMax)
x = -1*maxheap[0]
print(float(x))
else:
toMin = -1*hq.heappushpop(maxheap, -1*x)
hq.heappush(minheap, toMin)
x = (-1*maxheap[0]+minheap[0])/2.0
print(x)
```

## View More Similar Problems

## Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

View Solution →## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

View Solution →## Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

View Solution →## Direct Connections

Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do

View Solution →## Subsequence Weighting

A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =

View Solution →## Kindergarten Adventures

Meera teaches a class of n students, and every day in her classroom is an adventure. Today is drawing day! The students are sitting around a round table, and they are numbered from 1 to n in the clockwise direction. This means that the students are numbered 1, 2, 3, . . . , n-1, n, and students 1 and n are sitting next to each other. After letting the students draw for a certain period of ti

View Solution →