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";
}
|
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");
}
|
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")
|
C#
static void fun()
{
int i, j;
for (i = 1; i <= n; i++)
for (j = 1; j <= log(i); j++)
Console.Write("GeeksforGeeks");
}
|
Javascript
const fun()
{
let i, j;
for (i = 1; i <= n; i++)
for (j = 1; j <= Math.log(i); j++)
document.write("GeeksforGeeks");
}
|
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
.png)
