Change in Median of given array after deleting given elements

Last Updated : 24 Mar, 2023

Given two arrays arr1[] and arr2[]. The array arr1[] is sorted. The task is to print the change in the median after removing each element from array arr2[] one by one.

Note: The array arr2[] has only those elements that are present in array arr1[].

Examples:

Input: arr1[] = {2, 4, 6, 8, 10}, arr2[] = {4, 6}
Output: 1 1
Explanation:
Initially median is 6.
After removing 4, array becomes arr1[] = {2, 6, 8, 10}, median = 7, therefore the difference is 7 - 6 = 1.
After removing 6, array becomes arr1[] = {2, 8, 10}, median = 8, therefore the difference is 8 - 7 = 1.

Input: arr1[] = {1, 100, 250, 251}, arr2[] = {250, 1}
Output: -75 75.5
Explanation:
Initially median is 175.
After removing 250, array becomes arr1[] = {1, 100, 251}, median = 100, therefore the difference is 100 - 175 = -75.
After removing 1, array becomes arr1[] = {100, 251}, median = 175.5, therefore the difference is 175.5 - 100 = 75.5.

Approach: The idea is to traverse each element of the array arr2[] and remove each element from the array arr1[] and store the median of the array arr1[] after each removal of element in an array(say temp[]). Print the consecutive difference of the elements of the array to get change in median after removing elements from arr2[].

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the median change
// after removing elements from arr2[]
void medianChange(vector<int>& arr1,
                  vector<int>& arr2)
{
    int N = arr1.size();

    // To store the median
    vector<float> median;

    // Store the current median

    // If N is odd
    if (N & 1) {
        median
            .push_back(arr1[N / 2] * 1.0);
    }

    // If N is even
    else {
        median
            .push_back((arr1[N / 2]
                        + arr1[(N - 1) / 2])
                       / 2.0);
    }

    for (auto& x : arr2) {

        // Find the current element
        // in arr1
        auto it = find(arr1.begin(),
                       arr1.end(),
                       x);

        // Erase the element
        arr1.erase(it);

        // Decrement N
        N--;

        // Find the new median
        // and append

        // If N is odd
        if (N & 1) {
            median
                .push_back(arr1[N / 2] * 1.0);
        }

        // If N is even
        else {
            median
                .push_back((arr1[N / 2]
                            + arr1[(N - 1) / 2])
                           / 2.0);
        }
    }

    // Print the corresponding
    // difference of median
    for (int i = 0;
         i < median.size() - 1;
         i++) {
        cout << median[i + 1] - median[i]
             << ' ';
    }
}

// Driven Code
int main()
{
    // Given arrays
    vector<int> arr1 = { 2, 4, 6, 8, 10 };
    vector<int> arr2 = { 4, 6 };

    // Function Call
    medianChange(arr1, arr2);

    return 0;
}

Java

// Java program for the 
// above approach
import java.util.*;

class GFG{
    
// Function to find the median 
// change after removing elements 
// from arr2[]
public static void medianChange(List<Integer> arr1,
                                List<Integer> arr2)
{
    int N = arr1.size();
 
    // To store the median
    List<Integer> median = new ArrayList<>(); 
 
    // Store the current median
 
    // If N is odd
    if ((N & 1) != 0)
        median.add(arr1.get(N / 2) * 1);
 
    // If N is even
    else
        median.add((arr1.get(N / 2) + 
                    arr1.get((N - 1) / 2)) / 2);
 
    for(int x = 0; x < arr2.size(); x++)
    {
        
        // Find the current element
        // in arr1
        int it = arr1.indexOf(arr2.get(x));
 
        // Erase the element
        arr1.remove(it);
 
        // Decrement N
        N--;
 
        // Find the new median
        // and append
 
        // If N is odd
        if ((N & 1) != 0) 
        {
            median.add(arr1.get(N / 2) * 1);
        }
        
        // If N is even
        else 
        {
            median.add((arr1.get(N / 2) +
                        arr1.get((N - 1) / 2)) / 2);
        }
    }
 
    // Print the corresponding
    // difference of median
    for(int i = 0; i < median.size() - 1; i++)
    {
        System.out.print(median.get(i + 1) -
                         median.get(i) + " ");
    }
}

// Driver Code              
public static void main(String[] args)
{
    
    // Given arrays
    List<Integer> arr1  = new ArrayList<Integer>(){
        { add(2); add(4); add(6); add(8); add(10); } };
    List<Integer> arr2 = new ArrayList<Integer>(){
        { add(4); add(6); } };
 
    // Function Call
    medianChange(arr1, arr2);
}
}

// This code is contributed by divyesh072019

Python3

# Python3 program for the 
# above approach

# Function to find the median 
# change after removing elements 
# from arr2[]
def medianChange(arr1, arr2):

    N = len(arr1)

    # To store the median
    median = []

    # Store the current median

    # If N is odd
    if (N & 1):
        median.append(arr1[N // 2] * 1)

    # If N is even
    else:
        median.append((arr1[N // 2] + 
                       arr1[(N - 1) // 2]) // 2)

    for x in arr2:

        # Find the current 
        # element in arr1
        it = arr1.index(x)

        # Erase the element
        arr1.pop(it)

        # Decrement N
        N -= 1

        # Find the new median
        # and append

        # If N is odd
        if (N & 1):
            median.append(arr1[N // 2] * 1)

        # If N is even
        else:
            median.append((arr1[N // 2] + 
                           arr1[(N - 1) // 2]) // 2)

    # Print the corresponding
    # difference of median
    for i in range(len(median) - 1):
        print(median[i + 1] - median[i],
              end = ' ')

# Driver Code
if __name__ == "__main__":

    # Given arrays
    arr1 = [2, 4, 6, 
            8, 10]
    arr2 = [4, 6]

    # Function Call
    medianChange(arr1, arr2)

# This code is contributed by Chitranayal

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{
    
// Function to find the median change
// after removing elements from arr2[]
static void medianChange(List<int> arr1,
                         List<int> arr2)
{
    int N = arr1.Count;
 
    // To store the median
    List<double> median = new List<double>(); 
 
    // Store the current median
 
    // If N is odd
    if ((N & 1) != 0) 
    {
        median.Add(arr1[N / 2] * 1.0);
    }
 
    // If N is even
    else 
    {
        median.Add((arr1[N / 2] + 
              arr1[(N - 1) / 2]) / 2.0);
    }
 
    foreach(int x in arr2)
    {
        
        // Find the current element
        // in arr1
        int it = arr1.IndexOf(x);
 
        // Erase the element
        arr1.RemoveAt(it);
 
        // Decrement N
        N--;
 
        // Find the new median
        // and append
 
        // If N is odd
        if ((N & 1) != 0)
        {
            median.Add(arr1[N / 2] * 1.0);
        }
 
        // If N is even
        else 
        {
            median.Add((arr1[N / 2] +
                  arr1[(N - 1) / 2]) / 2.0);
        }
    }
 
    // Print the corresponding
    // difference of median
    for(int i = 0; i < median.Count - 1; i++)
    {
        Console.Write(median[i + 1] - 
                      median[i] + " ");
    }
}

// Driver Code
static void Main() 
{
    
    // Given arrays
    List<int> arr1 = new List<int>(
        new int[]{ 2, 4, 6, 8, 10 });
    List<int> arr2 = new List<int>(
        new int[]{ 4, 6 });
 
    // Function Call
    medianChange(arr1, arr2);
}
}

// This code is contributed by divyeshrabadiya07

JavaScript

<script>

// JavaScript program for the
// above approach

// Function to find the median
// change after removing elements
// from arr2[]
function medianChange(arr1,arr2)
{
    let N = arr1.length;
  
    // To store the median
    let median = [];
  
    // Store the current median
  
    // If N is odd
    if ((N & 1))
        median.push((arr1[(Math.floor(N / 2))] * 1));
  
    // If N is even
    else
        median.push(Math.floor((arr1[(Math.floor(N / 2))] +
        arr1[(Math.floor((N - 1) / 2))]) / 2));
  
    for(let x = 0; x < arr2.length; x++)
    {
         
        // Find the current element
        // in arr1
        let it = arr1.indexOf(arr2[x]);
        // Erase the element
        arr1.splice(it,1);
  
        // Decrement N
        N--;
  
        // Find the new median
        // and append
  
        // If N is odd
        if ((N & 1))
        {
            median.push(arr1[(Math.floor(N / 2))] * 1);
        }
         
        // If N is even
        else
        {
            median.push(Math.floor((arr1[(Math.floor(N / 2))] +
            arr1[(Math.floor((N - 1) / 2))]) / 2));
        }
    }
    // Print the corresponding
    // difference of median
    for(let i = 0; i < median.length - 1; i++)
    {
        document.write((median[i + 1] -
                         median[i]) + " ");
    }
}
// Driver Code

// Given arrays
let arr1 = [2, 4, 6,
            8, 10];
let arr2 = [4, 6];

// Function Call
medianChange(arr1, arr2)


// This code is contributed by avanitrachhadiya2155

</script>

Output:

1 1

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

Change in Median of given array after deleting given elements

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