Median of sliding window in an array | Set 2

Prerequisites: Policy based data structure, Sliding window technique.
Given an array of integer arr[] and an integer K, the task is to find the median of each window of size K starting from the left and moving towards the right by one position each time.
Examples:

Input: arr[] = {-1, 5, 13, 8, 2, 3, 3, 1}, K = 3 
Output: 5 8 8 3 3 3 
Explanation:
1st Window: {-1, 5, 13} Median = 5 
2nd Window: {5, 13, 8} Median = 8 
3rd Window: {13, 8, 2} Median = 8 
4th Window: {8, 2, 3} Median = 3 
5th Window: {2, 3, 3} Median = 3 
6th Window: {3, 3, 1} Median = 3

Input: arr[] = {-1, 5, 13, 8, 2, 3, 3, 1}, K = 4 
Output: 6.5 6.5 5.5 3.0 2.5 

Naive Approach:
The simplest approach to solve the problem is to traverse over every window of size K and sort the elements of the window and find the middle element. Print the middle element of every window as the median. 
Time Complexity: O(N*KlogK) 
Auxiliary Space: O(K)
Sorted Set Approach: Refer to Median of sliding window in an array to solve the problem using SortedSet.
Ordered Set Approach:
In this article, an approach to solving the problem using a Policy-based Ordered set data structure. 
Follow the steps below to solve the problem: 

• Insert the first window of size K in the Ordered_set( maintains a sorted order). Hence, the middle element of this Ordered set is the required median of the corresponding window. 
  
• The middle element can be obtained by the find_by_order() method in O(logN) computational complexity.
• Proceed to the following windows by remove the first element of the previous window and insert the new element. To remove any element from the set, find the order of the element in the Ordered_Set using order_by_key(), which fetches the result in O(logN) computational complexity, and erase() that element by searching its obtained order in the Ordered_Set using find_by_order() method. Now add the new element for the new window.  
• Repeat the above steps for each window and print the respective medians.  

Below is the implementation of the above approach.

CPP

// C++ Program to implement the
// above approach
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

// Policy based data structure
typedef tree<int, null_type,
             less_equal<int>, rb_tree_tag,
             tree_order_statistics_node_update>
    Ordered_set;

// Function to find and return the
// median of every window of size k
void findMedian(int arr[], int n,
                int k)
{

    Ordered_set s;

    for (int i = 0; i < k; i++)
        s.insert(arr[i]);

    if (k & 1) {

        // Value at index k/2
        // in sorted list.
        int ans = *s.find_by_order(k / 2);

        cout << ans << " ";

        for (int i = 0; i < n - k; i++) {

            // Erasing Element out of window.
            s.erase(s.find_by_order(
                s.order_of_key(
                    arr[i])));

            // Inserting newer element
            // to the window
            s.insert(arr[i + k]);

            // Value at index k/2 in
            // sorted list.
            ans = *s.find_by_order(k / 2);

            cout << ans << " ";
        }
        cout << endl;
    }
    else {

        // Getting the two middle
        // median of sorted list.
        float ans = ((float)*s.find_by_order(
                         (k + 1) / 2 - 1)
                     + (float)*s.find_by_order(k
                                               / 2))
                    / 2;

        printf("%.2f ", ans);

        for (int i = 0; i < n - k; i++) {
            s.erase(s.find_by_order(
                s.order_of_key(arr[i])));

            s.insert(arr[i + k]);

            ans = ((float)*s.find_by_order(
                       (k + 1) / 2 - 1)
                   + (float)*s.find_by_order(k
                                             / 2))
                  / 2;

            printf("%.2f ", ans);
        }
        cout << endl;
    }
}

// Driver Code
int main()
{
    int arr[] = { -1, 5, 13, 8, 2,
                  3, 3, 1 };
    int k = 3;

    int n = sizeof(arr)
            / sizeof(arr[0]);
    findMedian(arr, n, k);

    return 0;
}

Python3

# Python program to implement the
# above approach

# Policy based data structure
# in Python (using sortedcontainers package)
from sortedcontainers import SortedList

def findMedian(arr, n, k):

    s = SortedList()

    for i in range(k):
        s.add(arr[i])

    if k & 1:

        # Value at index k/2 in sorted list.
        ans = s[k//2]

        print(ans, end=' ')

        for i in range(n-k):

            # Erasing Element out of window.
            s.remove(arr[i])

            # Inserting newer element to the window
            s.add(arr[i+k])

            # Value at index k/2 in sorted list.
            ans = s[k//2]

            print(ans, end=' ')
        print()
    else:

        # Getting the two middle median of sorted list.
        ans = (s[(k+1)//2-1] + s[k//2]) / 2.0

        print("%.2f" % ans, end=' ')

        for i in range(n-k):
            s.remove(arr[i])

            s.add(arr[i+k])

            ans = (s[(k+1)//2-1] + s[k//2]) / 2.0

            print("%.2f" % ans, end=' ')
        print()

# Driver Code
if __name__ == '__main__':
    arr = [-1, 5, 13, 8, 2, 3, 3, 1]
    k = 3

    n = len(arr)
    findMedian(arr, n, k)

Java

import java.util.Scanner;
import java.util.TreeSet;

public class GFG {

    // Function to find and return the median of every window of size k
    public static void findMedian(int[] arr, int n, int k) {

        TreeSet<Integer> s = new TreeSet<>();

        for (int i = 0; i < k; i++) {
            s.add(arr[i]);
        }

        if (k % 2 == 1) {
            // Value at index k/2 in sorted list.
            int ans = s.toArray(new Integer[0])[k / 2];

            System.out.print(ans + " ");

            for (int i = 0; i < n - k; i++) {
                // Erasing Element out of window.
                s.remove(arr[i]);

                // Inserting newer element to the window
                s.add(arr[i + k]);

                // Value at index k/2 in sorted list.
                ans = s.toArray(new Integer[0])[k / 2];

                System.out.print(ans + " ");
            }
            System.out.println();
        } else {
            // Getting the two middle median of sorted list.
            float ans = ((float)s.toArray(new Integer[0])[(k + 1) / 2 - 1]
                    + (float)s.toArray(new Integer[0])[k / 2]) / 2;

            System.out.printf("%.2f ", ans);

            for (int i = 0; i < n - k; i++) {
                s.remove(arr[i]);
                s.add(arr[i + k]);

                ans = ((float)s.toArray(new Integer[0])[(k + 1) / 2 - 1]
                        + (float)s.toArray(new Integer[0])[k / 2]) / 2;

                System.out.printf("%.2f ", ans);
            }
            System.out.println();
        }
    }

    // Driver Code
    public static void main(String[] args) {

        int[] arr = {-1, 5, 13, 8, 2, 3, 3, 1};
        int k = 3;

        int n = arr.length;
        findMedian(arr, n, k);
    }
}
// This code is contributed by Shivam Tiwari

Javascript

function findMedian(arr, n, k) {
    // Create a TreeSet to store the current window of elements
    let s = new Set();

    // Add the first k elements to the TreeSet
    for (let i = 0; i < k; i++) {
        s.add(arr[i]);
    }

    if (k % 2 == 1) {
        // If k is odd, the median is the value at index k/2 in the sorted list
        let ans = Array.from(s)[(Math.floor(k / 2))];

        process.stdout.write(`${ans} `);

        for (let i = 0; i < n - k; i++) {
            // Remove the element that's no longer in the window
            s.delete(arr[i]);

            // Add the next element to the window
            s.add(arr[i + k]);

            // Calculate the new median
            ans = Array.from(s)[(Math.floor(k / 2))];

            process.stdout.write(`${ans} `);
        }
        console.log();
    } else {
        // If k is even, the median is the average of the two middle elements in the sorted list
        let ans = (Array.from(s)[(k + 1) / 2 - 1] + Array.from(s)[k / 2]) / 2;

        process.stdout.write(`${ans.toFixed(2)} `);

        for (let i = 0; i < n - k; i++) {
            s.delete(arr[i]);
            s.add(arr[i + k]);

            ans = (Array.from(s)[(k + 1) / 2 - 1] + Array.from(s)[k / 2]) / 2;

            process.stdout.write(`${ans.toFixed(2)} `);
        }
        console.log();
    }
}

// Driver Code
let arr = [-1, 5, 13, 8, 2, 3, 3, 1];
let k = 3;
let n = arr.length;
findMedian(arr, n, k);
// This code is contributed by Shivam Tiwari

C#

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Collections;
using System.IO;

    class GFG
    {
        // Function to find and return the median of every window of size k
        static void FindMedian(int[] arr, int n, int k)
        {
            SortedSet<int> set = new SortedSet<int>();

            for (int i = 0; i < k; i++)
                set.Add(arr[i]);

            if (k % 2 != 0)
            {
                // Value at index k/2
                // in sorted list.
                int ans = set.ElementAt((k / 2));

                Console.Write(ans + " ");

                for (int i = 0; i < n - k; i++)
                {

                    // Erasing Element out of window.
                    set.Remove(arr[i]);

                    // Inserting newer element
                    // to the window
                    set.Add(arr[i + k]);

                    // Value at index k/2 in
                    // sorted list.
                    ans = set.ElementAt((k / 2));

                    Console.Write(ans + " ");
                }

                Console.WriteLine();
            }
            else
            {

                // Getting the two middle
                // median of sorted list.
                double ans = ((double)set.ElementAt((k + 1) / 2 - 1) +
                            (double)set.ElementAt(k / 2)) / 2;

                Console.Write(Math.Round(ans, 2) + " ");

                for (int i = 0; i < n - k; i++)
                {
                    set.Remove(arr[i]);

                    set.Add(arr[i + k]);

                    ans = ((double)set.ElementAt((k + 1) / 2 - 1) +
                            (double)set.ElementAt(k / 2)) / 2;

                    Console.Write(Math.Round(ans, 2) + " ");
                }

                Console.WriteLine();
            }
        }

        // Driver code
        static void Main(string[] args)
        {
            int[] arr = { -1, 5, 13, 8, 2, 3, 3, 1 };
            int k = 3;

            int n = arr.Length;
            FindMedian(arr, n, k);
        }
    }
// This code is contributed By Shivam Tiwari

Output

5 8 8 3 3 3 

Time Complexity: O(NlogK) 
Auxiliary Space: O(K)