Largest k numbers in sorted order Given a set of n numbers in an arbitrary order
ID: 3794328 • Letter: L
Question
Largest k numbers in sorted order
Given a set of n numbers in an arbitrary order, we wish to find k smallest among them and output them insorted order. For each of the following approaches, write down pseudocode for an algorithm that implements the method mentioned and analyze your algorithm. Your analysis should be in terms of n and k. Your algorithm should exhibit the best asymptotic worst-case running time for that approach.
Sort the numbers in increasing order and output the first k.
Build a min-priority queue from the numbers, and call Extract-Min k times.
Use an order-statistic algorithm to find the k-th smallest number, partition around that number, and sort the k smallest numbers before returning them.
How do the above approaches compare for various values of k?
Explanation / Answer
#include<iostream>
#include<algorithm>
#include<climits>
using namespace std;
int partition(int arr[], int l, int r, int k);
// A simple function to find median of arr[]. This is called
// only for an array of size 5 in this program.
int findMedian(int arr[], int n)
{
sort(arr, arr+n); // Sort the array
return arr[n/2]; // Return middle element
}
// Returns k'th smallest element in arr[l..r] in worst case
// linear time. ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT
int kthSmallest(int arr[], int l, int r, int k)
{
// If k is smaller than number of elements in array
if (k > 0 && k <= r - l + 1)
{
int n = r-l+1; // Number of elements in arr[l..r]
// Divide arr[] in groups of size 5, calculate median
// of every group and store it in median[] array.
int i, median[(n+4)/5]; // There will be floor((n+4)/5) groups;
for (i=0; i<n/5; i++)
median[i] = findMedian(arr+l+i*5, 5);
if (i*5 < n) //For last group with less than 5 elements
{
median[i] = findMedian(arr+l+i*5, n%5);
i++;
}
// Find median of all medians using recursive call.
// If median[] has only one element, then no need
// of recursive call
int medOfMed = (i == 1)? median[i-1]:
kthSmallest(median, 0, i-1, i/2);
// Partition the array around a random element and
// get position of pivot element in sorted array
int pos = partition(arr, l, r, medOfMed);
// If position is same as k
if (pos-l == k-1)
return arr[pos];
if (pos-l > k-1) // If position is more, recur for left
return kthSmallest(arr, l, pos-1, k);
// Else recur for right subarray
return kthSmallest(arr, pos+1, r, k-pos+l-1);
}
// If k is more than number of elements in array
return INT_MAX;
}
void swap(int *a, int *b)
{
int temp = *a;
*a = *b;
*b = temp;
}
// It searches for x in arr[l..r], and partitions the array
// around x.
int partition(int arr[], int l, int r, int x)
{
// Search for x in arr[l..r] and move it to end
int i;
for (i=l; i<r; i++)
if (arr[i] == x)
break;
swap(&arr[i], &arr[r]);
// Standard partition algorithm
i = l;
for (int j = l; j <= r - 1; j++)
{
if (arr[j] <= x)
{
swap(&arr[i], &arr[j]);
i++;
}
}
swap(&arr[i], &arr[r]);
return i;
}
// Driver program to test above methods
int main()
{
int arr[] = {12, 3, 5, 7, 4, 19, 26};
int n = sizeof(arr)/sizeof(arr[0]), k = 3;
cout << "K'th smallest element is "
<< kthSmallest(arr, 0, n-1, k);
return 0;
}
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.