Program for Mean and median of an unsorted array

Given n size unsorted array, find it’s mean and median. 

Mean of an array = (sum of all elements) / (number of elements)

Median of a sorted array of size n is defined as the middle element when n is odd and average of middle two elements when n is even.
Since the array is not sorted here, we sort the array first, then apply above formula.

Examples:

Input  : a[] = {1, 3, 4, 2, 6, 5, 8, 7}
Output : Mean = 4.5
         Median = 4.5
Sum of the elements is 1 + 3 + 4 + 2 + 6 + 
5 + 8 + 7 = 36
Mean = 36/8 = 4.5
Since number of elements are even, median
is average of 4th and 5th largest elements.
which means (4 + 5)/2 = 4.5

Input  : a[] = {4, 4, 4, 4, 4}
Output : Mean = 4
         Median = 4 

Below is the code implementation: 

C++

// CPP program to find mean and median of
// an array
#include <bits/stdc++.h>
using namespace std;

// Function for calculating mean
double findMean(int a[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += a[i];

    return (double)sum / (double)n;
}

// Function for calculating median
double findMedian(int a[], int n)
{
    // First we sort the array
    sort(a, a + n);

    // check for even case
    if (n % 2 != 0)
        return (double)a[n / 2];

    return (double)(a[(n - 1) / 2] + a[n / 2]) / 2.0;
}

// Driver code
int main()
{
    int a[] = { 1, 3, 4, 2, 7, 5, 8, 6 };
    int n = sizeof(a) / sizeof(a[0]);

    // Function call
    cout << "Mean = " << findMean(a, n) << endl;
    cout << "Median = " << findMedian(a, n) << endl;
    return 0;
}

Java

// Java program to find mean
// and median of an array
import java.util.*;

class GFG
{
    // Function for calculating mean
    public static double findMean(int a[], int n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += a[i];

        return (double)sum / (double)n;
    }

    // Function for calculating median
    public static double findMedian(int a[], int n)
    {
        // First we sort the array
        Arrays.sort(a);

        // check for even case
        if (n % 2 != 0)
            return (double)a[n / 2];

        return (double)(a[(n - 1) / 2] + a[n / 2]) / 2.0;
    }

    // Driver code
    public static void main(String args[])
    {
        int a[] = { 1, 3, 4, 2, 7, 5, 8, 6 };
        int n = a.length;

        // Function call
        System.out.println("Mean = " + findMean(a, n));
        System.out.println("Median = " + findMedian(a, n));
    }
}

// This article is contributed by Anshika Goyal.

Python3

# Python3 program to find mean
# and median of an array

# Function for calculating mean


def findMean(a, n):

    sum = 0
    for i in range(0, n):
        sum += a[i]

    return float(sum/n)

# Function for calculating median


def findMedian(a, n):

    # First we sort the array
    sorted(a)

    # check for even case
    if n % 2 != 0:
        return float(a[int(n/2)])

    return float((a[int((n-1)/2)] +
                  a[int(n/2)])/2.0)


# Driver code
a = [1, 3, 4, 2, 7, 5, 8, 6]
n = len(a)

# Function call
print("Mean =", findMean(a, n))
print("Median =", findMedian(a, n))

# This code is contributed by Smitha Dinesh Semwal

C#

// C# program to find mean
// and median of an array
using System;

class GFG
{
    // Function for
    // calculating mean
    public static double findMean(int[] a, int n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += a[i];

        return (double)sum / (double)n;
    }

    // Function for
    // calculating median
    public static double findMedian(int[] a, int n)
    {
        // First we sort
        // the array
        Array.Sort(a);

        // check for
        // even case
        if (n % 2 != 0)
            return (double)a[n / 2];

        return (double)(a[(n - 1) / 2] + a[n / 2]) / 2.0;
    }

    // Driver Code
    public static void Main()
    {
        int[] a = { 1, 3, 4, 2, 7, 5, 8, 6 };
        int n = a.Length;

        // Function call
        Console.Write("Mean = " + findMean(a, n) + "\n");
        Console.Write("Median = " + findMedian(a, n)
                      + "\n");
    }
}

// This code is contributed by Smitha .

PHP

<?php
// PHP program to find mean
// and median of an array

// Function for calculating mean
function findMean(&$a, $n)
{
    $sum = 0;
    for ($i = 0; $i < $n; $i++)
        $sum += $a[$i];

    return (double)$sum /
           (double)$n;
}

// Function for
// calculating median
function findMedian(&$a, $n)
{
    // First we sort the array
    sort($a);

    // check for even case
    if ($n % 2 != 0)
    return (double)$a[$n / 2];

    return (double)($a[($n - 1) / 2] +
                    $a[$n / 2]) / 2.0;
}

// Driver Code
$a = array(1, 3, 4, 2,
           7, 5, 8, 6);
$n = sizeof($a);

// Function call
echo "Mean = " .
      findMean($a, $n)."\n";
echo "Median = " .
      findMedian($a, $n);

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>

// Javascript program to find mean
// and median of an array

// Function for
// calculating mean
function findMean(a,n)
{
    let sum = 0;
    for (let i = 0; i < n; i++)
        sum += a[i];

    return sum / n;
}

// Function for
// calculating median
function findMedian(a,n)
{
    // First we sort
    // the array
    a.sort();

    // check for
    // even case
    if (n % 2 != 0)
        return a[n / 2];

    return (a[Math.floor((n-1)/2)] +
            a[n / 2]) / 2;
}

// Driver Code

let a = [1, 3, 4, 2, 7, 5, 8, 6]
let n = a.length;

// Function call
document.write("Mean = " + findMean(a, n) + "<br>");
document.write("Median = " + findMedian(a, n));


</script>

Output

Mean = 4.5
Median = 4.5

Time Complexity to find mean: O(n) 
Time Complexity to find median: O(n Log n) as we need to sort the array first. Note that we can find median in O(n) time using methods

Auxiliary Space : O(1)

discussed here and here.

This article is contributed by Himanshu Ranjan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.