A Time Complexity Question

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++)

Time Complexity of the 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)) 

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.

Recommended Posts:

2.3 Average Difficulty : 2.3/5.0
Based on 67 vote(s)