Time complexity and number of elements operated on
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Algorithm A executes an operation of time complexity O(log n) on an array storing n elements.
Algorithm B chooses log n elements from an array, and performs an O(n) calculation on each.
Since d(n) = O(f(n)) and e(n) = O(g(n)), then d(n)*e(n) = O(f(n)*g(n)), does that mean that algorithms A and B both have a time complexity of O(n log n)?
algorithm time-complexity big-o
edited Nov 10 at 23:03
Scott Hunter
32.7k63970
asked Nov 10 at 22:55
erykkk
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up vote
0
down vote
favorite
Algorithm A executes an operation of time complexity O(log n) on an array storing n elements.
Algorithm B chooses log n elements from an array, and performs an O(n) calculation on each.
Since d(n) = O(f(n)) and e(n) = O(g(n)), then d(n)*e(n) = O(f(n)*g(n)), does that mean that algorithms A and B both have a time complexity of O(n log n)?
algorithm time-complexity big-o
edited Nov 10 at 23:03
Scott Hunter
32.7k63970
asked Nov 10 at 22:55
erykkk
466
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Algorithm A executes an operation of time complexity O(log n) on an array storing n elements.
Algorithm B chooses log n elements from an array, and performs an O(n) calculation on each.
Since d(n) = O(f(n)) and e(n) = O(g(n)), then d(n)*e(n) = O(f(n)*g(n)), does that mean that algorithms A and B both have a time complexity of O(n log n)?
algorithm time-complexity big-o
edited Nov 10 at 23:03
Scott Hunter
32.7k63970
asked Nov 10 at 22:55
erykkk
466
Algorithm A executes an operation of time complexity O(log n) on an array storing n elements.
Algorithm B chooses log n elements from an array, and performs an O(n) calculation on each.
Since d(n) = O(f(n)) and e(n) = O(g(n)), then d(n)*e(n) = O(f(n)*g(n)), does that mean that algorithms A and B both have a time complexity of O(n log n)?
algorithm time-complexity big-o
algorithm time-complexity big-o
edited Nov 10 at 23:03
Scott Hunter
32.7k63970
asked Nov 10 at 22:55
erykkk
466
edited Nov 10 at 23:03
Scott Hunter
32.7k63970
asked Nov 10 at 22:55
erykkk
466
edited Nov 10 at 23:03
Scott Hunter
32.7k63970
edited Nov 10 at 23:03
Scott Hunter
32.7k63970
edited Nov 10 at 23:03
Scott Hunter
32.7k63970
32.7k63970
asked Nov 10 at 22:55
erykkk
466
asked Nov 10 at 22:55
erykkk
466
asked Nov 10 at 22:55
erykkk
466
466
add a comment |
add a comment |
2 Answers
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The algorithm A has a time complexity of O(log n) and the algorithm B has a time complexity of O(n * log n). Algorithm B calculates something with O(n) on log * n elements. I assume that choosing is equivalent to sorting.
answered Nov 10 at 23:03
Charlie
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1
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Assuming that B doesn't take too long to choose the elements, and you meant Algorithm A does O(log n) work on each element, yes.
answered Nov 10 at 23:02
Scott Hunter
32.7k63970
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The algorithm A has a time complexity of O(log n) and the algorithm B has a time complexity of O(n * log n). Algorithm B calculates something with O(n) on log * n elements. I assume that choosing is equivalent to sorting.
answered Nov 10 at 23:03
Charlie
5442626
add a comment |
up vote
2
down vote
The algorithm A has a time complexity of O(log n) and the algorithm B has a time complexity of O(n * log n). Algorithm B calculates something with O(n) on log * n elements. I assume that choosing is equivalent to sorting.
answered Nov 10 at 23:03
Charlie
5442626
add a comment |
up vote
2
down vote
up vote
2
down vote
The algorithm A has a time complexity of O(log n) and the algorithm B has a time complexity of O(n * log n). Algorithm B calculates something with O(n) on log * n elements. I assume that choosing is equivalent to sorting.
answered Nov 10 at 23:03
Charlie
5442626
The algorithm A has a time complexity of O(log n) and the algorithm B has a time complexity of O(n * log n). Algorithm B calculates something with O(n) on log * n elements. I assume that choosing is equivalent to sorting.
answered Nov 10 at 23:03
Charlie
5442626
answered Nov 10 at 23:03
Charlie
5442626
answered Nov 10 at 23:03
Charlie
5442626
answered Nov 10 at 23:03
Charlie
5442626
5442626
add a comment |
add a comment |
up vote
1
down vote
Assuming that B doesn't take too long to choose the elements, and you meant Algorithm A does O(log n) work on each element, yes.
answered Nov 10 at 23:02
Scott Hunter
32.7k63970
add a comment |
up vote
1
down vote
Assuming that B doesn't take too long to choose the elements, and you meant Algorithm A does O(log n) work on each element, yes.
answered Nov 10 at 23:02
Scott Hunter
32.7k63970
add a comment |
up vote
1
down vote
up vote
1
down vote
Assuming that B doesn't take too long to choose the elements, and you meant Algorithm A does O(log n) work on each element, yes.
answered Nov 10 at 23:02
Scott Hunter
32.7k63970
Assuming that B doesn't take too long to choose the elements, and you meant Algorithm A does O(log n) work on each element, yes.
answered Nov 10 at 23:02
Scott Hunter
32.7k63970
answered Nov 10 at 23:02
Scott Hunter
32.7k63970
answered Nov 10 at 23:02
Scott Hunter
32.7k63970
answered Nov 10 at 23:02
Scott Hunter
32.7k63970
32.7k63970
add a comment |
add a comment |
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