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

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

- We need to find k numbers which have a minimum difference with the median of the data set.
- It includes sorting using quick sort and then printing k numbers as a result.
- The time complexity will be O(n*log(n)+k).

- Take input of the data.
- Sort the data using the quick-sort algorithm.
- Starting from the median index, using two variable moves towards the end of the array.
- compare each value to the median and print those which are closer to it.
- Repeat this k time printing the k numbers closest to the median.
- Exit.

#include<iostream>
 
using namespace std;
 
// Swapping two values.
void swap(int *a, int *b)
{
	int temp; 
	temp = *a;
	*a = *b;
	*b = temp;
}
 
// Partitioning the array on the basis of values at high as pivot value.
int Partition(int a[], int low, int high)
{
	int pivot, index, i;
	index = low;
	pivot = high;
 
	// Getting index of pivot.
	for(i=low; i < high; i++)
	{
		if(a[i] < a[pivot])
		{
			swap(&a[i], &a[index]);
			index++;
		}
	}
	// Swapping value at high and at the index obtained.
	swap(&a[pivot], &a[index]);
 
	return index;
}
 
// Implementing QuickSort algorithm.
int QuickSort(int a[], int low, int high)
{
	int pindex;
	if(low < high)
	{
		// Partitioning array using randomized pivot.
		pindex = Partition(a, low, high);
		// Recursively implementing QuickSort.
		QuickSort(a, low, pindex-1);
		QuickSort(a, pindex+1, high);
	}
	return 0;
}
 
int main()
{
	int n, i, high, low, k;
	double d1,d2, median;
	cout<<"Enter the number of element in dataset: ";
	cin>>n;
 
	int a[n];
	// Taking input of the data set.
	for(i = 0; i < n; i++)
	{
		cout<<"\nEnter "<<i+1<<"th element: ";
		cin>>a[i];
	}
 
	cout<<"\nEnter the number of element nearest to the median required: ";
	cin>>k;
 
	// Sort the data.
	QuickSort(a, 0, n-1);
 
	//Print the result.
	cout<<"\tThe K element nearest to the median are: ";
	// Check the number of data element to be even or odd and proceed accordingly.
	if(n%2 == 1)
	{
		median = a[n/2];
		high = n/2+1;
		low = n/2;
 
		// Loop to search for the next element generating minimum difference from median.
		while(k > 0)
		{
			// If difference from the first half element is minimum.
			if((median-a[low] <= a[high]-median) && low >= 0)
			{
				cout<<" "<<a[low];
				low--;
				k--;
			}
			// If difference from the Second half element is minimum.
			else if((median-a[low] > a[high]-median) && high <= n-1)
			{
				cout<<" "<<a[high];
				high++;
				k--;
			}
		}
	}
	else
	{
		// Need to use floating variable since the median can be in the fractional form.
		d1 = a[n/2];
		d2 = a[n/2-1];
		median = (d1+d2)/2;
		high = n/2;
		low = n/2-1;
		while(k > 0)
		{
			d1 = a[low];
			d2 = a[high];
			// If difference from the first half element is minimum.
			if((median-d2 <= d1-median) && low >= 0)
			{
				cout<<" "<<a[low];
				low--;
				k--;
			}
			// If difference from the Second half element is minimum.
			else if((median-d2 > d1-median) && high <= n-1)
			{
				cout<<" "<<a[high];
				high++;
				k--;
			}
		}
	}
 
	return 0;
}