An array is $d$-sorted if every key in the array is located at a distance at most $d$ from its location in the sorted array.

I need to write an algorithm that get a $d$-sorted array of length $n$ and sort it in the following running times:

  1. $\Theta(n)$ if $d$ is constant
  2. $\Theta(n\log\log n)$ if $d = \Theta(\log n)$.

My attempt:

I wrote the following pseudo-code:

 min-heap-size <- d
 for i <- 1 to n 
      if min-heap not empty
        then EXTRACT-MIN <- A[i]
          if i+d<=n
            then heap-insert-min(min-heap,A[i+d])   

But in terms of runtime, all I get is $O(n\log\log d)$.

My method: I initialize $i \gets 1$ and then I build a min-heap that contains the first $d$ elements.

As long as the heap is not empty, I use EXTRACT-MIN and put the element at index $i$ in the array. If $i+d \le n$, then I add the element at index $i+d$ to the min-heap.

Any help?

  • 1
    $\begingroup$ An algorithm with running time $O(n \log \log d)$ would allow to sort $n$ elements in time $O(n \log \log n)$ (just pick $d=n$). Since your algorithm is comparison based, this contradicts the lower bound of $\Omega(n \log n)$ on the running time of any comparison-based sorting algorithm. $\endgroup$
    – Steven
    Apr 12, 2021 at 15:11
  • $\begingroup$ Your algorithm probably runs in time $O(n\log d)$, which satisfies your two constraints. $\endgroup$ Apr 12, 2021 at 16:59

1 Answer 1


Since your heap contains $d$ elements, every operation takes take $O(\log d)$, for a total running time of $O(n\log d)$.

When $d$ is constant, this is $O(n)$, and when $d = O(\log n)$, this is $O(n\log\log n)$.

For a matching $\Omega(n\log d)$ lower bound, see this question.


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