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A Time Complexity Question
  • Difficulty Level : Easy
  • Last Updated : 29 Oct, 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++)

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(n*log(n)) 

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

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