How to find k nearest neighbors to the median of n distinct numbers in O(n) time?

Question

I can use the median of medians selection algorithm to find the median in O(n). Also, I know that after the algorithm is done, all the elements to the left of the median are les

Answer 1

All the answers suggesting to subtract the median from the array would produce incorrect results. This method will find the elements closest in value, not closest in position.

For example, if the array is 1,2,3,4,5,10,20,30,40. For k=2, the value returned would be (3,4); which is incorrect. The correct output should be (4,10) as they are the nearest neighbor.

The correct way to find the result would be using the selection algorithm to find upper and lower bound elements. Then by direct comparison find the remaining elements from the list.

Answer 2

med=Select(A,1,n,n/2)   //finds the median

for i=1 to n
   B[i]=mod(A[i]-med)

q=Select(B,1,n,k) //get the kth smallest difference

j=0
for i=1 to n
   if B[i]<=q 
     C[j]=A[i] //A[i], the real value should be assigned instead of B[i] which is only the difference between A[i] and median.
       j++
return C

Answer 3

No one seems to quite have this. Here's how to do it. First, find the median as described above. This is O(n). Now park the median at the end of the array, and subtract the median from every other element. Now find element k of the array (not including the last element), using the quick select algorithm again. This not only finds element k (in order), it also leaves the array so that the lowest k numbers are at the beginning of the array. These are the k closest to the median, once you add the median back in.

Answer 4

You could use a non-comparison sort, such as a radix sort, on the list of numbers L, then find the k closest neighbors by considering windows of k elements and examining the window endpoints. Another way of stating "find the window" is find i that minimizes abs(L[(n-k)/2+i] - L[n/2]) + abs(L[(n+k)/2+i] - L[n/2]) (if k is odd) or abs(L[(n-k)/2+i] - L[n/2]) + abs(L[(n+k)/2+i+1] - L[n/2]) (if k is even). Combining the cases, abs(L[(n-k)/2+i] - L[n/2]) + abs(L[(n+k)/2+i+!(k&1)] - L[n/2]). A simple, O(k) way of finding the minimum is to start with i=0, then slide to the left or right, but you should be able to find the minimum in O(log(k)).

The expression you minimize comes from transforming L into another list, M, by taking the difference of each element from the median.

m=L[n/2]
M=abs(L-m)

i minimizes M[n/2-k/2+i] + M[n/2+k/2+i].

Answer 5

First select the median in O(n) time, using a standard algorithm of that complexity. Then run through the list again, selecting the elements that are nearest to the median (by storing the best known candidates and comparing new values against these candidates, just like one would search for a maximum element).

In each step of this additional run through the list O(k) steps are needed, and since k is constant this is O(1). So the total for time needed for the additional run is O(n), as is the total runtime of the full algorithm.

Answer 6

You can solve your problem like that:

You can find the median in O(n), w.g. using the O(n) nth_element algorithm.

You loop through all elements substutiting each with a pair:

the absolute difference to the median, element's value. 

Once more you do nth_element with n = k. after applying this algorithm you are guaranteed to have the k smallest elements in absolute difference first in the new array. You take their indices and DONE!