# A Time Complexity Question

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))