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Big O
I got this question , it is a theory question on Big 0
1-if n is 2 what is the final value for counter
2-if n is 5 what is the final value for counter
3-in light of this derive a expression for the number of times the inner loop will run in terms of n
4- state the Big O complexity
int counter =0
for (int i = n; i <=3*n; i ++){
while(i >+n){
i --:
counter ++
}
}
can anyone tell me how to get the values for counter
Related threads:
Re: Big O
What happens when you compile and execute the code and print out the results?
Please edit your post and wrap your code with
[code=java]
<YOUR CODE HERE>
[/code]
to get highlighting and preserve formatting.
If you don't understand my answer, don't ignore it, ask a question.
Re: Big O
it is not a question about the code, it is a theory question on Big O complexity. It asks to figure out what the value for counter is when the values are put in for n.
Re: Big O
Have you tried executing the code? what does it do?
If you don't understand my answer, don't ignore it, ask a question.
Re: Big O
program keeps running
Re: Big O
Then it has infinite complexity.Originally Posted by shanem
If you suspect that it is intended to finish eventually then post the code you are using. And describe the program's behaviour: especially the output suggested by Norm in #2.
[Edit] I think a couple of values for n as suggested by the original question is insufficient to see anything much. Try getting the values of counter for a whole bunch of values of n - say from zero to ten - and then graphing them (counter vs n).
Re: Big O
The question is not on the code , it is not meant to be used to run a program, it is a theory question to figure out the value for COUNTER when you put the values in for firstly n=2 and then n=5
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