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Home/ Questions/Q 9153083
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T12:12:11+00:00 2026-06-17T12:12:11+00:00

What time-complexity will the following code have in respect to the parameter size? Motivate.

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What time-complexity will the following code have in respect to the parameter size? Motivate.

// Process(A, N) is O(sqrt(N)). 

Function Complex(array[], size){
    if(size == 1) return 1;
    if(rand() / float(RAND_MAX) < 0.1){
        return Process(array, size*size)
             + Complex(array, size/2)
             + Process(array, size*size);
    }
}

I think it is O(N), because if Process(A, N) is O(sqrt(N)), then Process(A, N*N) should be O(N), and Complex(array, size/2) is O(log(n)) because it halves the size every time it runs. So on one run it takes O(N) + O(log(N)) + O(N) = O(N).

Please correct me and give me some hints on how I should think / proceed an assignment like this.

I appreciate all help and thanks in advance.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T12:12:12+00:00Added an answer on June 17, 2026 at 12:12 pm

    The time complexity of the algorithm is O(N) indeed, but for a different reason.

    The complexity of the function can be denoted as T(n) where:

    T(n) = T(n/2)       +       2*n
            ^                   ^
          recursive          2 calls to 
          invokation        Process(arr,n*n),
                              each is O(n(
    

    This recursion is well known to be O(n):

    T(n) = T(n/2) + 2*n = 
         = T(n/4) + 2*n/2 + 2*n = 
         = T(n/8) + 2*n/4 + 2*n/2 + 2*n
         = ....
         = 2*n / (2^logN) + ... + 2*n/2 + 2*n
         < 4n
         in O(n)
    

    Let’s formally prove it, we will use mathematical induction for it:

    Base: T(1) < 4 (check)

    Hypothesis: For n, and for every k<n the claim T(k) < 4k holds true.

    For n:

    T(n) = T(n/2) + n*2 = (*) 
         < 2*n + 2*n 
         = 4n
    

    Conclusion: T(n) is in O(n)

    (*) From the induction hypothesis

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