**Find the Median**

### Problem Statement :

The median of a list of numbers is essentially its middle element after sorting. The same number of elements occur after it as before. Given a list of numbers with an odd number of elements, find the median? Function Description Complete the findMedian function in the editor below. findMedian has the following parameter(s): int arr[n]: an unsorted array of integers Returns int: the median of the array Input Format The first line contains the integer n, the size of arr. The second line contains n space-separated integers arr[i]

### Solution :

` ````
Solution in C :
In C++ :
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int n, *arr;
cin>>n;
arr = new int[n];
for (int i = 0 ; i < n ; i++) cin>>arr[i];
sort(arr,arr+n);
if (n%2 == 1) cout<<arr[(n-1)/2]<<endl;
else cout<<(arr[n/2 - 1]+arr[n/2])/2<<endl;
return 0;
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static int quickSort(int[] a, int p, int r,int index)
{
if(p==r)
return a[p];
if(p<r)
{
int q=partition(a,p,r);
if(q==index)
return a[q];
else if(q>index)
return quickSort(a,p,q-1,index);
else
return quickSort(a,q+1,r,index);
}
return 0;
}
private static int partition(int[] a, int p, int r) {
int x = a[r];
int i = p-1 ;
int j = p ;
for(;j<r;++j)
{
if(a[j]<=x)
{
++i;
int temp = a[i];
a[i]=a[j];
a[j]=temp;
}
}
a[r]=a[i+1];
a[i+1]=x;
return i+1;
}
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int[] arr = new int[N];
for(int i=0;i<N;++i)
arr[i]=sc.nextInt();
System.out.println(quickSort(arr, 0, N-1,N/2));
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <assert.h>
int compare (const void * a, const void * b)
{
return ( *(int*)a - *(int*)b );
}
/* Tail starts here */
int main()
{
int _ar_size;
scanf("%d", &_ar_size);
int _ar[_ar_size], _ar_i;
for(_ar_i = 0; _ar_i < _ar_size; _ar_i++)
{
scanf("%d", &_ar[_ar_i]);
}
qsort (_ar, _ar_size, sizeof(int), compare);
printf("%d",_ar[_ar_size/2]);
return 0;
}
In Python3 :
import math
def main():
testcase = int(input())
med = [int(i) for i in input().split()]
print(findmed(med))
def findmed(med):
med.sort()
if(len(med)%2 == 1):
return med[int(len(med)/2)]
else:
return (med[math.floor(len(med)/2)+math.ceil(len(med)/2)])/2
main()
```

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