Maximize the median of an array

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++

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// 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;
}

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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

Output:

4 3 3 7 8 1 0 4


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Improved By : rituraj_jain