Median of Stream of Running Integers using STL | Set 2
Last Updated :
17 Mar, 2023
Given an array arr[] of size N representing integers required to be read as a data stream, the task is to calculate and print the median after reading every integer.
Examples:
Input: arr[] = { 5, 10, 15 } Output: 5 7.5 10 Explanation: After reading arr[0] from the data stream, the median is 5. After reading arr[1] from the data stream, the median is 7.5. After reading arr[2] from the data stream, the median is 10.
Input: arr[] = { 1, 2, 3, 4 } Output: 1 1.5 2 2.5
Approach: The problem can be solved using Ordered Set. Follow the steps below to solve the problem:
- Initialize a multi Ordered Set say, mst to store the array elements in a sorted order.
- Traverse the array using variable i. For every ith element insert arr[i] into mst and check if the variable i is even or not. If found to be true then print the median using (*mst.find_by_order(i / 2)).
- Otherwise, print the median by taking the average of (*mst.find_by_order(i / 2)) and (*mst.find_by_order((i + 1) / 2)).
Below is the implementation of the above approach:
C++
// C++ program to implement
// the above approach
#include <iostream>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type,
less_equal<int>, rb_tree_tag,
tree_order_statistics_node_update> idxmst;
// Function to find the median
// of running integers
void findMedian(int arr[], int N)
{
// Initialise a multi ordered set
// to store the array elements
// in sorted order
idxmst mst;
// Traverse the array
for (int i = 0; i < N; i++) {
// Insert arr[i] into mst
mst.insert(arr[i]);
// If i is an odd number
if (i % 2 != 0) {
// Stores the first middle
// element of mst
double res
= *mst.find_by_order(i / 2);
// Stores the second middle
// element of mst
double res1
= *mst.find_by_order(
(i + 1) / 2);
cout<< (res + res1) / 2.0<<" ";
}
else {
// Stores middle element of mst
double res
= *mst.find_by_order(i / 2);
// Print median
cout << res << " ";
}
}
}
// Driver Code
int main()
{
// Given stream of integers
int arr[] = { 1, 2, 3, 3, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function call
findMedian(arr, N);
}
# Python program to implement the approach for finding the median of running integers
# Import the necessary module for Ordered Dict
from collections import OrderedDict
def find_median(arr):
# Initialize an ordered dictionary to store the elements in sorted order
ordered_dict = OrderedDict()
# Traverse the array
for i in range(len(arr)):
# Insert arr[i] into ordered_dict
ordered_dict[arr[i]] = ordered_dict.get(arr[i], 0) + 1
# If i is an odd number
if i % 2 != 0:
# Find the middle elements and store them in a list
mid = list(ordered_dict.keys())[i//2:i//2 + 2]
# Calculate the median by taking the average of the middle elements
median = (mid[0] + mid[1]) / 2
# Print median
print("%.1f" % median, end=" ")
else:
# Find the middle element
mid = list(ordered_dict.keys())[i//2]
# Print median
print(mid, end=" ")
# Given stream of integers
arr = [1, 2, 3, 3, 4]
# Function call
find_median(arr)
# This code is contributed by Shivam Tiwari
// JavaScript program to implement the approach for finding the median of running integers
// Initialize an object to store the elements in sorted order
let orderedObj = {};
function find_median(arr) {
// Traverse the array
for (let i = 0; i < arr.length; i++) {
// Insert arr[i] into orderedObj
orderedObj[arr[i]] = (orderedObj[arr[i]] || 0) + 1;
// If i is an odd number
if (i % 2 !== 0) {
// Find the middle elements and store them in a list
let mid = Object.keys(orderedObj).slice(i / 2, i / 2 + 2);
// Calculate the median by taking the average of the middle elements
let median = (parseInt(mid[0]) + parseInt(mid[1])) / 2;
// Print median
process.stdout.write(median.toFixed(1) + " ");
} else {
// Find the middle element
let mid = Object.keys(orderedObj)[i / 2];
// Print median
process.stdout.write(mid + " ");
}
}
}
// Given stream of integers
let arr = [1, 2, 3, 3, 4];
// Function call
find_median(arr);
// This code is contributed by sdeadityasharma
import java.util.*;
public class GFG {
// Function to find the median
// of running integers
public static void findMedian(int[] arr) {
// Initialize an ordered dictionary to store the elements in sorted order
Map<Integer, Integer> ordered_dict = new TreeMap<>();
// Traverse the array
for (int i = 0; i < arr.length; i++) {
// Insert arr[i] into ordered_dict
ordered_dict.put(arr[i], ordered_dict.getOrDefault(arr[i], 0) + 1);
// If i is an odd number
if (i % 2 != 0) {
// Find the middle elements and store them in a list
List<Integer> mid = new ArrayList<>(ordered_dict.keySet()).subList(i / 2, i / 2 + 2);
// Calculate the median by taking the average of the middle elements
double median = (mid.get(0) + mid.get(1)) / 2.0;
// Print median
System.out.print(String.format("%.1f", median) + " ");
} else {
// Find the middle element
int mid = new ArrayList<>(ordered_dict.keySet()).get(i / 2);
// Print median
System.out.print(mid + " ");
}
}
}
// Driver Code
public static void main(String[] args) {
// Given stream of integers
int[] arr = {1, 2, 3, 3, 4};
// Function call
findMedian(arr);
}
}
// This code is contributed By Shivam Tiwari
//C# program to implement the approach for finding the median of running integers
using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
static void Main(string[] args)
{
// Given stream of integers
int[] arr = { 1, 2, 3, 3, 4 };
// Function call
find_median(arr);
}
// Function to find the median of running integers
static void find_median(int[] arr)
{
// Initialize an ordered dictionary to store the elements in sorted order
Dictionary<int, int> ordered_dict = new Dictionary<int, int>();
// Traverse the array
for (int i = 0; i < arr.Length; i++)
{
// Insert arr[i] into ordered_dict
if (ordered_dict.ContainsKey(arr[i]))
{
ordered_dict[arr[i]]++;
}
else
{
ordered_dict[arr[i]] = 1;
}
// If i is an odd number
if (i % 2 != 0)
{
// Find the middle elements and store them in a list
var mid = ordered_dict.Keys.ToList().GetRange(i / 2, 2);
// Calculate the median by taking the average of the middle elements
var median = (mid[0] + mid[1]) / 2.0;
// Print median
Console.Write("{0:F1} ", median);
}
else
{
// Find the middle element
var mid = ordered_dict.Keys.ToList().GetRange(i / 2, 1);
// Print median
Console.Write("{0} ", mid[0]);
}
}
}
}
// This code is contributed by shivamsharma215
Time Complexity: O(N * log(N))
Auxiliary Space: O(N)
Similar Reads
Median of Stream of Running Integers using STL Given that integers are being read from a data stream. Find the median of all the elements read so far starting from the first integer till the last integer. This is also called the Median of Running Integers. The data stream can be any source of data, for example, a file, an array of integers, inpu
9 min read
Median Of Running Stream of Numbers - (using Set) Given that integers are being read from a data stream. Find the median of all the elements read so far starting from the first integer until the last integer. This is also called Median of Running Integers. The given link already contains solution of this problem using Priority Queue. However, the f
9 min read
Mode in a stream of integers (running integers) Given that integers are being read from a data stream. Find the mode of all the elements read so far starting from the first integer till the last integer. Mode is defined as the element which occurs the maximum time. If two or more elements have the same maximum frequency, then take the one with th
5 min read
Sort a stream of integers Given an array, arr[] of size N whose elements from left to right, must be read as an incoming stream of integers, the task is to sort the stream of integers and print accordingly. Examples: Input: arr[] = {2, 4, 1, 7, 3}Output: 1 2 3 4 7Explanation: First element in the stream: 2 ? Sorted list: {2}
12 min read
Find Median from Running Data Stream Given a data stream arr[] where integers are read sequentially, the task is to determine the median of the elements encountered so far after each new integer is read.There are two cases for median on the basis of data set size.If the data set has an odd number then the middle one will be consider as
14 min read