Constant factor of an array Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm...
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Constant factor of an array
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
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In Elements of Programming Interviews in Python by Aziz, Lee and Prakash, they state on page 41:
Insertion into a full array can be handled by resizing, i.e.,
allocating a new array with additional memory and copying over the
entries from the original array. This increases the worst-case time of
insertion, but if the new array has, for example, a constant factor
larger than the original array, the average time for insertion is
constant since resizing is infrequent.
I grasp the concept of amortization that seems to be implied here, yet they seem to imply that in other cases, a newly allocated array could possess a constant factor smaller than the original array. Is that so? What does "constant factor" mean in this particular context? I'm having trouble understanding what's being said here.
algorithm-analysis
asked 2 hours ago
lispHK01lispHK01
1061
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$begingroup$
In Elements of Programming Interviews in Python by Aziz, Lee and Prakash, they state on page 41:
Insertion into a full array can be handled by resizing, i.e.,
allocating a new array with additional memory and copying over the
entries from the original array. This increases the worst-case time of
insertion, but if the new array has, for example, a constant factor
larger than the original array, the average time for insertion is
constant since resizing is infrequent.
I grasp the concept of amortization that seems to be implied here, yet they seem to imply that in other cases, a newly allocated array could possess a constant factor smaller than the original array. Is that so? What does "constant factor" mean in this particular context? I'm having trouble understanding what's being said here.
algorithm-analysis
asked 2 hours ago
lispHK01lispHK01
1061
New contributor
lispHK01 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In Elements of Programming Interviews in Python by Aziz, Lee and Prakash, they state on page 41:
Insertion into a full array can be handled by resizing, i.e.,
allocating a new array with additional memory and copying over the
entries from the original array. This increases the worst-case time of
insertion, but if the new array has, for example, a constant factor
larger than the original array, the average time for insertion is
constant since resizing is infrequent.
I grasp the concept of amortization that seems to be implied here, yet they seem to imply that in other cases, a newly allocated array could possess a constant factor smaller than the original array. Is that so? What does "constant factor" mean in this particular context? I'm having trouble understanding what's being said here.
algorithm-analysis
asked 2 hours ago
lispHK01lispHK01
1061
New contributor
lispHK01 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In Elements of Programming Interviews in Python by Aziz, Lee and Prakash, they state on page 41:
Insertion into a full array can be handled by resizing, i.e.,
allocating a new array with additional memory and copying over the
entries from the original array. This increases the worst-case time of
insertion, but if the new array has, for example, a constant factor
larger than the original array, the average time for insertion is
constant since resizing is infrequent.
I grasp the concept of amortization that seems to be implied here, yet they seem to imply that in other cases, a newly allocated array could possess a constant factor smaller than the original array. Is that so? What does "constant factor" mean in this particular context? I'm having trouble understanding what's being said here.
algorithm-analysis
algorithm-analysis
asked 2 hours ago
lispHK01lispHK01
1061
New contributor
lispHK01 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
lispHK01lispHK01
1061
New contributor
lispHK01 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
lispHK01lispHK01
1061
New contributor
lispHK01 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
lispHK01lispHK01
1061
asked 2 hours ago
lispHK01lispHK01
1061
1061
New contributor
lispHK01 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
lispHK01 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
lispHK01 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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That's probably a typo or poor wording -- in the quote, "has" should be "is".
answered 2 hours ago
D.W.♦D.W.
104k14130296
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
That's probably a typo or poor wording -- in the quote, "has" should be "is".
answered 2 hours ago
D.W.♦D.W.
104k14130296
$endgroup$
add a comment |
$begingroup$
That's probably a typo or poor wording -- in the quote, "has" should be "is".
answered 2 hours ago
D.W.♦D.W.
104k14130296
$endgroup$
add a comment |
$begingroup$
That's probably a typo or poor wording -- in the quote, "has" should be "is".
answered 2 hours ago
D.W.♦D.W.
104k14130296
$endgroup$
That's probably a typo or poor wording -- in the quote, "has" should be "is".
answered 2 hours ago
D.W.♦D.W.
104k14130296
answered 2 hours ago
D.W.♦D.W.
104k14130296
answered 2 hours ago
D.W.♦D.W.
104k14130296
answered 2 hours ago
D.W.♦D.W.
104k14130296
104k14130296
add a comment |
add a comment |
lispHK01 is a new contributor. Be nice, and check out our Code of Conduct.
lispHK01 is a new contributor. Be nice, and check out our Code of Conduct.
lispHK01 is a new contributor. Be nice, and check out our Code of Conduct.
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