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 << "w3wiki" ;
}
|
C
void fun()
{
int i, j;
for (i = 1; i <= n; i++)
for (j = 1; j <= log (i); j++)
printf ( "w3wiki" );
}
|
Java
static void fun()
{
int i, j;
for (i = 1 ; i <= n; i++)
for (j = 1 ; j <= log(i); j++)
System.out.printf( "w3wiki" );
}
|
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 ( "w3wiki" )
|
C#
static void fun()
{
int i, j;
for (i = 1; i <= n; i++)
for (j = 1; j <= log(i); j++)
Console.Write( "w3wiki" );
}
|
Javascript
const fun()
{
let i, j;
for (i = 1; i <= n; i++)
for (j = 1; j <= Math.log(i); j++)
document.write( "w3wiki" );
}
|
Time Complexity of the above function can be written as ?(log 1) + ?(log 2) + ?(log 3) + . . . . + ?(log n) which is ?(log n!)
Order of growth of ‘log n!’ and ‘n log n’ is same for large values of n, i.e., ?(log n!) = ?(n log n). So time complexity of fun() is ?(n log n).
The expression ?(log n!) = ?(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))
Sources:
http://en.wikipedia.org/wiki/Stirling%27s_approximation