Constant factor of an array












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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.










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    1












    $begingroup$


    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.










    share|cite|improve this question







    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.







    $endgroup$















      1












      1








      1





      $begingroup$


      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.










      share|cite|improve this question







      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






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      asked 2 hours ago









      lispHK01lispHK01

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          That's probably a typo or poor wording -- in the quote, "has" should be "is".






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            $begingroup$

            That's probably a typo or poor wording -- in the quote, "has" should be "is".






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              $begingroup$

              That's probably a typo or poor wording -- in the quote, "has" should be "is".






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                $begingroup$

                That's probably a typo or poor wording -- in the quote, "has" should be "is".






                share|cite|improve this answer









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                That's probably a typo or poor wording -- in the quote, "has" should be "is".







                share|cite|improve this answer












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                answered 2 hours ago









                D.W.D.W.

                104k14130296




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