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Home/ Questions/Q 8983047
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T20:44:12+00:00 2026-06-15T20:44:12+00:00

I am working on a problem that has a sorted array of n elements

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I am working on a problem that has a sorted array of n elements followed by an unsorted array of length

  1. O(logn)
  2. O(sqrt(n))

How to sort the entire list most efficiently? Which sorting should I use in the above two cases?

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  1. Editorial Team
    Editorial Team
    2026-06-15T20:44:13+00:00Added an answer on June 15, 2026 at 8:44 pm

    Since inserting a single element into array and keeping it sorted is O(n), you cannot get better then that.

    Thus, for both cases – sorting the smaller array and then using merge(part1,part2) will be O(n), and thus optimal in terms of asymptotic complexity.

    • sorting the smaller array: O(logn*loglog(n)), or O(sqrt(n)*log(sqrt(n)) respectively of the cases.
    • merge(part1,part2): O(n+logn) or O(n+sqrt(n)), which is O(n)1 anyway.

    So, the total complexity of both cases is O(n), which is optimal for this problem.


    (1) It is true because, log(n)^k is asymptotically smaller then n^m for each k>0,m>0, and specifically for k=1, m=1/2.

    Proof is based on taking logs on both sides:

    log (log(n)^k) <? log(n^m) <=>
    k*log(log(n)) <? m*log(n)
    

    The last is obviously true (for large n and constant k,m>0), and thus the claim is true.

    From this we can conclude that sqrt(n)*log(n) < sqrt(n) * n^1/2 = n, and thus it is indeed O(n).

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