Given an unsorted array a[] of size N, the task is to find its mean and median.
Mean of an array = (sum of all elements) / (number of elements)
The 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
Explanation: 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 Median = (4 + 5)/2 = 4.5Input: a[] = {4, 4, 4, 4, 4}
Output: Mean = 4, Median = 4
Approach: To solve the problem follow the below steps:
To find median:
- First, simply sort the array
- Then, check if the number of elements present in the array is even or odd
- If odd, then simply return the mid value of the array
- Else, the median is the average of the two middle values
To find Mean:
- At first, find the sum of all the numbers present in the array.
- Then, simply divide the resulted sum by the size of the array
Below is the code implementation:
C
// C program to find mean and median of// an array#include <stdio.h>#include <stdlib.h>// Function to compare two integers for qsortint cmpfunc(const void* a, const void* b){ return (*(int*)a - *(int*)b);}// Function for calculating meandouble 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 mediandouble findMedian(int a[], int n){ // First we sort the array qsort(a, n, sizeof(int), cmpfunc); // 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 codeint main(){ int a[] = { 1, 3, 4, 2, 7, 5, 8, 6 }; int N = sizeof(a) / sizeof(a[0]); // Function call printf("Mean = %f\n", findMean(a, N)); printf("Median = %f\n", findMedian(a, N)); return 0;} |
C++
// CPP program to find mean and median of// an array#include <bits/stdc++.h>using namespace std;// Function for calculating meandouble 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 mediandouble 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 codeint 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 arrayimport 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 meandef findMean(a, n): sum = 0 for i in range(0, n): sum += a[i] return float(sum/n)# Function for calculating mediandef 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 codea = [1, 3, 4, 2, 7, 5, 8, 6]n = len(a)# Function callprint("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 arrayusing 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 meanfunction findMean(&$a, $n){ $sum = 0; for ($i = 0; $i < $n; $i++) $sum += $a[$i]; return (double)$sum / (double)$n;}// Function for // calculating medianfunction 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($a);// Function callecho "Mean = " . findMean($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 meanfunction findMean(a,n){ let sum = 0; for (let i = 0; i < n; i++) sum += a[i]; return sum / n;}// Function for// calculating medianfunction 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 Codelet a = [1, 3, 4, 2, 7, 5, 8, 6]let n = a.length;// Function calldocument.write("Mean = " + findMean(a, n) + "<br>");document.write("Median = " + findMedian(a, n));</script> |
Output
Mean = 4.5 Median = 4.5
Complexity Analysis:
Time Complexity to find mean: O(N)
Time Complexity to find median: O(N Log N) as we need to sort the array first.
Auxiliary Space: O(1)
This article is contributed by Himanshu Ranjan. If you like neveropen 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 neveropen main page and help other Geeks.
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