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)?










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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)?










    share|improve this question


























      up vote
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      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)?










      share|improve this question















      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






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      edited Nov 10 at 23:03









      Scott Hunter

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      asked Nov 10 at 22:55









      erykkk

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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.






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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.






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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.






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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.






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                  up vote
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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.






                  share|improve this answer












                  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.







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                  answered Nov 10 at 23:03









                  Charlie

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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.






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                        up vote
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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.






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                          up vote
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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.






                          share|improve this answer












                          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.







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                          answered Nov 10 at 23:02









                          Scott Hunter

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                          32.7k63970






























                               

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