is O(long n!)
n! is O(n^n) by sterling approximation
hence it is O(log n^n) = O(nlogn)
On Sun, Oct 28, 2012 at 11:14 PM, Pralay Biswas
pralaybiswas2...@gmail.comwrote:
@ Siddharth :
Well, here is how you may understand it:
1) There is an outer loop that runs n times. (k)
2) Then there is an
Its nlogn dude
On Wed, Oct 31, 2012 at 8:03 PM, jai gupta jaigupt...@ymail.com wrote:
is O(long n!)
n! is O(n^n) by sterling approximation
hence it is O(log n^n) = O(nlogn)
On Sun, Oct 28, 2012 at 11:14 PM, Pralay Biswas
pralaybiswas2...@gmail.com wrote:
@ Siddharth :
Well, here is
Can somebody explain how it is O(n log n).
What is the significance of while loop in the above code?
I understand that the for loop implies O(n),does the log n in the O(n log
n) comes from the while loop?
What if there where two while loops in the for loop separately?
On Sat, Oct 27, 2012 at
while loop : logj base 2 ..
== log1 + log2 + ... logn = log(n!) [since logab = loga + logb]
== O(log(n^n)) = O(nlogn)
On Sun, Oct 28, 2012 at 3:56 AM, Siddharth Malhotra
codemalho...@gmail.comwrote:
Can somebody explain how it is O(n log n).
What is the significance of while loop in the above
@ Siddharth :
Well, here is how you may understand it:
1) There is an outer loop that runs n times. (k)
2) Then there is an inner loop where j is initially set to current k, but
it halves itself in every iteration
-- So for example, if k is 32, and j halves every time, then the
inner
O(log( n! ));
On Sat, Oct 27, 2012 at 6:03 AM, payal gupta gpt.pa...@gmail.com wrote:
That 's correct.
On Sat, Oct 27, 2012 at 3:25 AM, rahul sharma rahul23111...@gmail.comwrote:
for k=1 to n
{
j=k;
while(j0)
j=j/2;
}
the complexity is big o is o(nlogn)
am i ryt
--
You
Yes, its complexity is O(nlogn).
On Sat, Oct 27, 2012 at 2:35 PM, CHIRANJEEV KUMAR
cse.chiranj...@gmail.comwrote:
O(log( n! ));
On Sat, Oct 27, 2012 at 6:03 AM, payal gupta gpt.pa...@gmail.com wrote:
That 's correct.
On Sat, Oct 27, 2012 at 3:25 AM, rahul sharma
Yes!
On Fri, Oct 26, 2012 at 2:55 PM, rahul sharma rahul23111...@gmail.comwrote:
for k=1 to n
{
j=k;
while(j0)
j=j/2;
}
the complexity is big o is o(nlogn)
am i ryt
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for k=1 to n
{
j=k;
while(j0)
j=j/2;
}
the complexity is big o is o(nlogn)
am i ryt
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