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An interesting time complexity question

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  • Difficulty Level : Easy
  • Last Updated : 30 Oct, 2015
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What is the time complexity of following function fun()?




int fun(int n)
{    
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j < n; j += i)
        {
            // Some O(1) task
        }
    }    
}

For i = 1, the inner loop is executed n times.
For i = 2, the inner loop is executed approximately n/2 times.
For i = 3, the inner loop is executed approximately n/3 times.
For i = 4, the inner loop is executed approximately n/4 times.
…………………………………………………….
…………………………………………………….
For i = n, the inner loop is executed approximately n/n times.

So the total time complexity of the above algorithm is (n + n/2 + n/3 + … + n/n)

Which becomes n * (1/1 + 1/2 + 1/3 + … + 1/n)

The important thing about series (1/1 + 1/2 + 1/3 + … + 1/n) is, it is equal to Θ(Logn) (See this for reference). So the time complexity of the above code is Θ(nLogn).

As a side note, the sum of infinite harmonic series is counterintuitive as the series diverges. The value of timecomplex is ∞. This is unlike geometric series as geometric series with ratio less than 1 converges.

Reference:
http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29#Rate_of_divergence
http://staff.ustc.edu.cn/~csli/graduate/algorithms/book6/chap03.htm

This article is contributed by Rahul. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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