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A Time Complexity Question

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  • Difficulty Level : Easy
  • Last Updated : 27 Dec, 2021
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What is the time complexity of following function fun()? Assume that log(x) returns log value in base 2. 

C++




void fun()
{
    int i, j;
    for (i = 1; i <= n; i++)
        for (j = 1; j <= log(i); j++)
            cout << "GeeksforGeeks";
}
 
// This code is contributed by SHUBHAMSINGH10.

C




void fun()
{
    int i, j;
    for (i = 1; i <= n; i++)
        for (j = 1; j <= log(i); j++)
            printf("GeeksforGeeks");
}

Java




static void fun()
{
    int i, j;
    for (i = 1; i <= n; i++)
        for (j = 1; j <= log(i); j++)
            System.out.printf("GeeksforGeeks");
}
 
// This code is contributed by umadevi9616

Python3




import math
def fun():
    i = 0
    j = 0
    for i in range(1, n + 1):
        for j in range(1,math.log(i) + 1):
            print("GeeksforGeeks")
 
# This code is contributed by SHUBHAMSINGH10.

C#




static void fun()
{
    int i, j;
    for (i = 1; i <= n; i++)
        for (j = 1; j <= log(i); j++)
            Console.Write("GeeksforGeeks");
}
 
// This code is contributed by umadevi9616

Javascript




const fun()
{
    let i, j;
    for (i = 1; i <= n; i++)
        for (j = 1; j <= Math.log(i); j++)
            document.write("GeeksforGeeks");
}
 
// This code is contributed by SHUBHAMSINGH10.

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.
Sources: 
http://en.wikipedia.org/wiki/Stirling%27s_approximation


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