GATE | Sudo GATE 2020 Mock II (10 January 2019) | Question 39
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++)
printf("GeeksforGeeks");
}
(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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Last Updated :
09 Jan, 2020
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