# C++ Program to Find k Numbers Closest to Median of S, Where S is a Set of n Numbers

C++Server Side ProgrammingProgramming

This is a C++ Program to find k numbers closest to Median of S, where S is a set of n numbers.

## Algorithms

Begin
function partition() for partitioning the array on the basis of values at high as pivot value:
Arguments:
a[]=an array.
l=low
H=high
Body of the function:
Declare variables pivot, in, i
Initialize in = l
Set pivot = h
For i=l to h-1
if(a[i] < a[pivot])
swap a[i] and a[in])
increment in.
swap a[pivot] and a[in]
return in.
End
Begin
function QuickSort() to implement quicksort algorithm to sort the data elements:
Arguments:
a[]=an array.
l=low
h=high
Body of the function:
declare pindex
if(l < h)
index = Partition(a, l, h)
QuickSort(a, l, pindex-1);
QuickSort(a, pindex+1, h);
Return 0.
End
Begin
Function main(),
If the number of the data element are odd,
Assign the middle index to low and the index next to it to high and calculate median.
Run a loop for k times and print the element which his closer to the median.
Else
The median will be an average of two middle values .
Run a loop for k times and print the element which his closer to the median.
End

## Example

#include<iostream>
using namespace std;
void swap(int *x, int *y) { //swapping two values
int tmp;
tmp = *x;
*x = *y;
*y = tmp;
}
int Partition(int a[], int l, int h) {
int pivot, in, i;
in = l;
pivot = h;
for(i=l; i < h; i++) {
if(a[i] < a[pivot]) {
swap(&a[i], &a[in]);
in++;
}
}
swap(&a[pivot], &a[in]);
return in;
}
int QuickSort(int a[], int l, int h) {
int pindex;
if(l < h) {
pindex = Partition(a, l, h);
QuickSort(a, l, pindex-1);
QuickSort(a, pindex+1, h);
}
return 0;
}
int main() {
int n, i, h, l, k;
double d1,d2, median;
cout<<"Enter the number of element in dataset: ";
cin>>n;
int a[n];
for(i = 0; i < n; i++) {
cout<<"\nEnter "<<i+1<<" element: ";
cin>>a[i];
}
cout<<"\nEnter the number of element nearest to the median required: ";
cin>>k;
QuickSort(a, 0, n-1);
cout<<"The K element nearest to the median are: ";
if(n%2 == 1) {
median = a[n/2];
h = n/2+1;
l= n/2;
while(k > 0) {
if((median-a[l] <= a[h]-median) && l >= 0) {
cout<<" "<<a[l];
l--;
k--;
} else if((median-a[l] > a[h]-median) && h <= n-1) {
cout<<" "<<a[h];
h++;
k--;
}
}
} else {
d1 = a[n/2];
d2 = a[n/2-1];
median = (d1+d2)/2;
h = n/2;
l = n/2-1;
while(k > 0) {
d1 = a[l];
d2 = a[h];
if((median-d2 <= d1-median) && l >= 0) {
cout<<" "<<a[l];
l--;
k--;
} else if((median-d2 > d1-median) && h <= n-1) {
cout<<" "<<a[h];
h++;
k--;
}
}
}
return 0;
}

## Output

Enter the number of element in dataset: 7
Enter 1 element: 7
Enter 2 element: 6
Enter 3 element: 5
Enter 4 element: 4
Enter 5 element: 3
Enter 6 element: 2
Enter 7 element: 1
Enter the number of element nearest to the median required: 2
The K element nearest to the median are: 4 3