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Maximize the median of an array

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  • Difficulty Level : Basic
  • Last Updated : 30 Jun, 2022

Given an array of n elements. The task is to print the array in a form such that the median of that array is maximum.

Examples: 

Input: arr[] = {3, 1, 2, 3, 8}
Output: 3 1 8 2 3

Input: arr[] = {9, 8, 7, 6, 5, 4}
Output: 7 6 9 8 5 4

Approach: Median of any sequence is the middle most element of given array. Hence if the array has an odd number of elements the n/2 th element is the median of the array and in the case, if the array has even number of elements, then n/2th and n/2 – 1 th element are median.
To maximize the median of any array, first of all, check whether its size is even or odd depending upon the size of array perform following steps. 

  1. If size is odd: Find the maximum element from array and swap it with the n/2th element.
  2. If size is even: Find the first two maximum element and swap them with n/2th and n/2-1 th elements.

Below is the implementation of the above approach: 

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print array with maximum median
void printMaxMedian(int arr[], int n)
{
    // case when size of array is odd
    if (n % 2) {
        int* maxElement = max_element(arr, arr + n);
        swap(*maxElement, arr[n / 2]);
    }
 
    // when the size of the array is even
    else {
        // find 1st maximum element
        int* maxElement1 = max_element(arr, arr + n);
 
        // find 2nd maximum element
        int* maxElement2 = max_element(arr, maxElement1);
        maxElement2 = max(maxElement2,
                          max_element(maxElement1 + 1, arr + n));
 
        // swap position for median
        swap(*maxElement1, arr[n / 2]);
        swap(*maxElement2, arr[n / 2 - 1]);
    }
 
    // print resultant array
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
 
// Driver code
int main()
{
    int arr[] = { 4, 8, 3, 1, 3, 7, 0, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printMaxMedian(arr, n);
 
    return 0;
}


Java




// Java implementation of the above approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG{
 
public static int getIndexOfLargest(int[] array,
                                    int st,
                                    int n)
{
    if (array == null || array.length == 0)
     
        // null or empty
        return -1;
 
    int largest = st;
    for(int i = st + 1; i < n; i++)
    {
        if (array[i] > array[largest])
            largest = i;
    }
     
    // Position of the first largest
    // found
    return largest;
}
 
public static void swap(int a, int b, int[] arr)
{
    int temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp;
}
 
// Function to print array with maximum median
public static void printMaxMedian(int[] arr, int n)
{
     
    // Case when size of array is odd
    if (n % 2 != 0)
    {
        int maxElement = getIndexOfLargest(
            arr, 0, arr.length);
             
        swap(maxElement, n / 2, arr);
    }
 
    // When the size of the array is even
    else
    {
         
        // Find 1st maximum element
        int maxElement1 = getIndexOfLargest(
            arr, 0, arr.length);
 
        // Find 2nd maximum element
        int maxElement2 = getIndexOfLargest(
            arr, 0, maxElement1);
             
        maxElement2 = Math.max(
            arr[maxElement2],
            getIndexOfLargest(arr, maxElement1 + 1,
                              arr.length));
 
        // Swap position for median
        swap(maxElement1, n / 2, arr);
        swap(maxElement2, n / 2 - 1, arr);
    }
 
    // Print resultant array
    for(int i = 0; i < n; i++)
        System.out.print(arr[i] + " ");
}
 
// Driver Code
public static void main(String[] args)
    throws java.lang.Exception
{
    int arr[] = { 4, 8, 3, 1, 3, 7, 0, 4 };
 
    printMaxMedian(arr, arr.length);
}
}
 
// This code is contributed by grand_master


Python3




# Python3 implementation of the above approach
 
# Function to print the array with
# maximum median
def printMaxMedian(arr, n):
 
    # case when size of the array is odd
    if n % 2 != 0:
        maxElement = arr.index(max(arr))
        (arr[maxElement],
         arr[n // 2]) = (arr[n // 2],
                         arr[maxElement])
 
    # when size of array is even
    else:
         
        # find 1st maximum element
        maxElement1 = arr.index(max(arr))
 
        # find 2nd maximum element
        maxElement2 = arr.index(max(arr[0 : maxElement1]))
        maxElement2 = arr.index(max(arr[maxElement2],
                                max(arr[maxElement1 + 1:])))
 
        # swap position for median
        (arr[maxElement1],
         arr[n // 2]) = (arr[n // 2],
                         arr[maxElement1])
        (arr[maxElement2],
         arr[n // 2 - 1]) = (arr[n // 2 - 1],
                             arr[maxElement2])
 
    # print the resultant array
    for i in range(0, n):
        print(arr[i], end = " ")
 
# Driver code
if __name__ == "__main__":
 
    arr = [4, 8, 3, 1, 3, 7, 0, 4]
    n = len(arr)
    printMaxMedian(arr, n)
 
# This code is contributed by Rituraj Jain


C#




// C# implementation of the above approach
using System;
 
class GFG{
     
static int getIndexOfLargest(int[] array,
                             int st, int n)
{
    if (array == null || array.Length == 0)
     
        // Null or empty
        return -1;
  
    int largest = st;
    for(int i = st + 1; i < n; i++)
    {
        if (array[i] > array[largest])
            largest = i;
    }
      
    // Position of the first largest
    // found
    return largest;
}
  
static void swap(int a, int b, int[] arr)
{
    int temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp;
}
  
// Function to print array with maximum median
static void printMaxMedian(int[] arr, int n)
{
     
    // Case when size of array is odd
    if (n % 2 != 0)
    {
        int maxElement = getIndexOfLargest(
            arr, 0, arr.Length);
              
        swap(maxElement, n / 2, arr);
    }
  
    // When the size of the array is even
    else
    {
         
        // Find 1st maximum element
        int maxElement1 = getIndexOfLargest(
            arr, 0, arr.Length);
  
        // Find 2nd maximum element
        int maxElement2 = getIndexOfLargest(
            arr, 0, maxElement1);
              
        maxElement2 = Math.Max(
            arr[maxElement2],
            getIndexOfLargest(arr, maxElement1 + 1,
                              arr.Length));
  
        // Swap position for median
        swap(maxElement1, n / 2, arr);
        swap(maxElement2, n / 2 - 1, arr);
    }
  
    // Print resultant array
    for(int i = 0; i < n; i++)
        Console.Write(arr[i] + " ");
}
 
// Driver Code
static void Main()
{
    int[] arr = { 4, 8, 3, 1, 3, 7, 0, 4 };
     
    printMaxMedian(arr, arr.Length);
}
}
 
// This code is contributed by divyeshrabadiya07


Output: 

4 3 3 7 8 1 0 4

 

Time Complexity: O(n), where n is the size of the given array
Auxiliary Space: O(1), as no extra space is used


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