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Asked: 2026-06-13T17:25:26+00:00

I’m pretty new to this stuff and I need your help. I should build

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I’m pretty new to this stuff and I need your help.

I should build an efficient simple algorithm which returns the maximum value in an array with size of n which contains the numbers 1,2,…n with repetitions.

Then I have to determine the best running time, average running time and worst running time.

So I have two questions:

  1. First of all I’m trying to understand what’s the idea of requiring an efficient solution for this simple algorithm. As far as I understand I should only have a simple loop from 1 to n and look for the maximum value. Is the “efficient” algorithm points out that If I find the value n in the array I can stop searching more values because this is the maximum value in the array?

  2. How do I determine the best running time and average running time using the fact, while calculating the average running time, that It’s a uniform distribution.
    That is, each cell in the array has a chance of 1/n to be the maximum value.

Thanks a lot in advance!

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1 Answer

  1. Editorial Team

    Best case – finding the max element as the first (O(1)), worst case – it is the last element checked (O(n)).

    The tricky part is the average case.

    To find the average case – we need the expected number of iterations!

    Since you can stop after you find the maximum, we can split the problem into two parts:

    1. [0,n-1): Since on average (assuming uniform independent distribution for each element) – the number n has probability 1/n to be in each place, then the expected number of iterations for this part is 1/n + 2*((n-1)/n)/n + 3 * ((n-1)/n)^2/n + ... + (n-1) * ((n-1)/n)^(n-2)/n 1

      The above formula yields an ugly formula which is O(n)
    2. The last element is going to be checked if the first n-1 elements did not contain the value n: so you need to add to the above n* ((n-1)/n)^(n-1), which is O(n) as well (lim to infinity is 1/e * n).

    This totals in O(n) average time solution.

    (1): The formula for each element is j*((n-1)/n)^(j-1) * (1/n) because:

    • j – for the number of elements checked (number of iterations)
    • ((n-1)/n)^(j-1) – Probability that there is no n in the previous elements
    • (1/n) – Probability this number is n.
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