I need to find a sorting algorithm to sort an array where the the $k$-th element appears $2^{k-1}$ times in $O(2^k)$.

It also given there are $k$ distinct elements in the array, hence there are $1 + 2 + 2^2 + \dots + 2^{k-1}=2^k-1$ elements in the array. So I actually need to sort the array in $O(n)$ where $n$ is the array size.

Not sure how to use the known quantity of each element to sort faster.

Any help is appreciated.

  • $\begingroup$ Can you give an example of such a list you want to sort? I dont understand if the $k$'th element is considered the $k$'th largest number, or the $k$'th element is literally the $k$'th new element you see when you scan the array from left to right. For example, is $[4,2,2,5,5,5,5,1,1,1,1,1,1,1,1]$ an example of such an array, or is $[4,4,4,4,2,2,5,5,5,5,5,5,5,5,1]$ an example of such an array? $\endgroup$
    – nir shahar
    Apr 25, 2021 at 9:38

1 Answer 1


If I understand correctly the title (which is a bit clearer than the post itself), an example of such an array would be $[1, 4, 1, 2, 1, 2, 5, 2, 1, 1, 4, 1, 2, 1, 1]$.

If that's the case, you can:

  • Count the number of occurrences of each value, with a hashtable for example;
  • Sort the unique values with descending order of occurrences;
  • Recreate the sorted array by repeating each value by the number of occurrences.

This algorithm is indeed in time complexity $O(n)$, since the array contains only $\log_2 (n+1)$ unique values. That means that even the worst case of operations in the hashtable is in complexity $O((\log n)^2)$, and sorting the unique values array can be done in $O(\log n \log \log n)$. The last step is done in $O(n)$.


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