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GATE | Sudo GATE 2020 Mock II (10 January 2019) | Question 39

  • Last Updated : 09 Jan, 2020

What is the time complexity of following function fun()? Assume that log(x) returns log value in base 2.

void fun()
   int i, j;
   for (i=1; i<=n; i++)
      for (j=1; j<=log(i); j++)

(A) Θ(n)
(B) Θ(nLogn)
(C) Θ(n^2)
(D) Θ(n^2(Logn))

Answer: (B)

Explanation: The time complexity of above function can be written as: Θ(log 1) + Θ(log 2) + Θ(log 3) + . . . . + Θ(log n) which is Θ (log n!)
Order of growth of ‘log n!’ and ‘n log n’ is same for large values of n, i.e., Θ (log n!) = Θ(n log n). So time complexity of fun() is Θ(n log n).

The expression Θ(log n!) = Θ(n log n) can be easily derived from following Stirling’s approximation (or Stirling’s formula)

log n! = n log n - n + O(log(n))

Option (B) is correct.

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