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)
                        




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