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What would the big-oh run time be for this loop?
for (int k =j/2; k < n/2; k++)
4 Replies - 1016 Views - Last Post: 25 October 2015 - 01:05 PM
#1
blvckphvroh
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Algorithm run time analysis
Posted 22 October 2015 - 10:54 PM
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#2
macosxnerd101
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Re: Algorithm run time analysis
Posted 22 October 2015 - 11:47 PM
Moved to Computer Science.
What have you tried? What do you think?
What have you tried? What do you think?
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blvckphvroh
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Re: Algorithm run time analysis
Posted 22 October 2015 - 11:53 PM
macosxnerd101, on 22 October 2015 - 11:47 PM, said:
Moved to Computer Science.
What have you tried? What do you think?
What have you tried? What do you think?
At first I was thinking it would just be O(n) but the divide by 2 threw me off and I started thinking it could be O(log n)
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macosxnerd101
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Re: Algorithm run time analysis
Posted 23 October 2015 - 10:57 AM
Note that O(log n) is a subset of O(n). However, O(n) is the correct bound. You start at some constant less than n and go to some constant multiple of n. If j == 0, it's still O(n), right?
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#5
ishkabible
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Re: Algorithm run time analysis
Posted 25 October 2015 - 01:05 PM
Is more context provided? What is j? How does it relate to n?
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