Median of Stream of Running Integers using STL

Median of Stream of Running Integers using STL | Set 2

Difficulty Level : Hard
Last Updated : 08 Dec, 2020

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 i^th 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); 
} 

Time Complexity: O(N * log(N))
Auxiliary Space: O(N)

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