Finding the maximum of a bitonic sequence












2












$begingroup$


Given to us is an array $B[1],ldots,B[n]$ as input, which satisfies the following property: there exists a special index $i^* ∈ {1, dots , n}$ such that $$B[1] < B[2] < dots < B[i^*− 1] < B[i^*] > B[i^* + 1] > B[i^* + 2] > dots > B[n].$$



It is noted that the index $i^*$ is not given as part of the input.



The question asks me to design an algorithm for computing the index $i^*$, whilst trying to ensure it has the smallest running time possible.



I just wondered if anyone could produce this algorithm and explain your thought process on how you came up with it if possible? Stuck on these type of questions






algorithms arrays discrete-mathematics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Has been asked before. Use some sort of binary search.
    $endgroup$
    – Yuval Filmus
    Feb 7 at 3:16










  • $begingroup$
    Could you link me to the question and yes, thanks
    $endgroup$
    – opp333
    Feb 7 at 3:48










  • $begingroup$
    stackoverflow.com/questions/39185197/…
    $endgroup$
    – Yuval Filmus
    Feb 7 at 4:18
















2












$begingroup$


Given to us is an array $B[1],ldots,B[n]$ as input, which satisfies the following property: there exists a special index $i^* ∈ {1, dots , n}$ such that $$B[1] < B[2] < dots < B[i^*− 1] < B[i^*] > B[i^* + 1] > B[i^* + 2] > dots > B[n].$$



It is noted that the index $i^*$ is not given as part of the input.



The question asks me to design an algorithm for computing the index $i^*$, whilst trying to ensure it has the smallest running time possible.



I just wondered if anyone could produce this algorithm and explain your thought process on how you came up with it if possible? Stuck on these type of questions






algorithms arrays discrete-mathematics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Has been asked before. Use some sort of binary search.
    $endgroup$
    – Yuval Filmus
    Feb 7 at 3:16










  • $begingroup$
    Could you link me to the question and yes, thanks
    $endgroup$
    – opp333
    Feb 7 at 3:48










  • $begingroup$
    stackoverflow.com/questions/39185197/…
    $endgroup$
    – Yuval Filmus
    Feb 7 at 4:18














2












2








2


1



$begingroup$


Given to us is an array $B[1],ldots,B[n]$ as input, which satisfies the following property: there exists a special index $i^* ∈ {1, dots , n}$ such that $$B[1] < B[2] < dots < B[i^*− 1] < B[i^*] > B[i^* + 1] > B[i^* + 2] > dots > B[n].$$



It is noted that the index $i^*$ is not given as part of the input.



The question asks me to design an algorithm for computing the index $i^*$, whilst trying to ensure it has the smallest running time possible.



I just wondered if anyone could produce this algorithm and explain your thought process on how you came up with it if possible? Stuck on these type of questions






algorithms arrays discrete-mathematics







share|cite|improve this question











$endgroup$




Given to us is an array $B[1],ldots,B[n]$ as input, which satisfies the following property: there exists a special index $i^* ∈ {1, dots , n}$ such that $$B[1] < B[2] < dots < B[i^*− 1] < B[i^*] > B[i^* + 1] > B[i^* + 2] > dots > B[n].$$



It is noted that the index $i^*$ is not given as part of the input.



The question asks me to design an algorithm for computing the index $i^*$, whilst trying to ensure it has the smallest running time possible.



I just wondered if anyone could produce this algorithm and explain your thought process on how you came up with it if possible? Stuck on these type of questions






algorithms arrays discrete-mathematics




algorithms arrays discrete-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 7 at 4:46









Yuval Filmus

195k14184349




195k14184349










asked Feb 7 at 3:01









opp333opp333

134




134








  • 1




    $begingroup$
    Has been asked before. Use some sort of binary search.
    $endgroup$
    – Yuval Filmus
    Feb 7 at 3:16










  • $begingroup$
    Could you link me to the question and yes, thanks
    $endgroup$
    – opp333
    Feb 7 at 3:48










  • $begingroup$
    stackoverflow.com/questions/39185197/…
    $endgroup$
    – Yuval Filmus
    Feb 7 at 4:18














  • 1




    $begingroup$
    Has been asked before. Use some sort of binary search.
    $endgroup$
    – Yuval Filmus
    Feb 7 at 3:16










  • $begingroup$
    Could you link me to the question and yes, thanks
    $endgroup$
    – opp333
    Feb 7 at 3:48










  • $begingroup$
    stackoverflow.com/questions/39185197/…
    $endgroup$
    – Yuval Filmus
    Feb 7 at 4:18








1




1




$begingroup$
Has been asked before. Use some sort of binary search.
$endgroup$
– Yuval Filmus
Feb 7 at 3:16




$begingroup$
Has been asked before. Use some sort of binary search.
$endgroup$
– Yuval Filmus
Feb 7 at 3:16












$begingroup$
Could you link me to the question and yes, thanks
$endgroup$
– opp333
Feb 7 at 3:48




$begingroup$
Could you link me to the question and yes, thanks
$endgroup$
– opp333
Feb 7 at 3:48












$begingroup$
stackoverflow.com/questions/39185197/…
$endgroup$
– Yuval Filmus
Feb 7 at 4:18




$begingroup$
stackoverflow.com/questions/39185197/…
$endgroup$
– Yuval Filmus
Feb 7 at 4:18










2 Answers
2






active

oldest

votes


















3












$begingroup$

The first thing that comes to my mind after reading your question is as follow. Draw the sequence as a graph given below. Each $B[i]$ you can think of a point in the graph given below.
enter image description here



The idea to do binary search should be clear from the question as input sequence is almost sorted. Now if you guess the peak point then you are in a good position, but during the binary search if you choose a number which will give you two cases.



enter image description here



The above two cases are the cases in which you need to think, but if you have tried few problems on binary search then now the problem will be easy to solve. More formally you can see the yuval's answer. You ask for the thought process, I have tried to give you my thought process. For any two persons thought process is different, there may be many other way to visualise the problem you have described.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    A sequence satisfying your constraint is known as bitonic. (Strictly speaking, a bitonic sequence is one which is either monotone increasing then decreasing or monotone decreasing then increasing). You can find the index $i^*$ using binary search. The idea is that given an index $i$, you can compare $i$ to $i^*$ as follows:




    • If $B[i-1] < B[i] > B[i+1]$, then $i = i^*$.

    • If $B[i-1] < B[i] < B[i+1]$, then $i < i^*$.

    • If $B[i-1] > B[i] > B[i+1]$, then $i > i^*$.


    I'll leave you to handle the corner cases $i = 1$ and $i = n$.






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      The first thing that comes to my mind after reading your question is as follow. Draw the sequence as a graph given below. Each $B[i]$ you can think of a point in the graph given below.
      enter image description here



      The idea to do binary search should be clear from the question as input sequence is almost sorted. Now if you guess the peak point then you are in a good position, but during the binary search if you choose a number which will give you two cases.



      enter image description here



      The above two cases are the cases in which you need to think, but if you have tried few problems on binary search then now the problem will be easy to solve. More formally you can see the yuval's answer. You ask for the thought process, I have tried to give you my thought process. For any two persons thought process is different, there may be many other way to visualise the problem you have described.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        The first thing that comes to my mind after reading your question is as follow. Draw the sequence as a graph given below. Each $B[i]$ you can think of a point in the graph given below.
        enter image description here



        The idea to do binary search should be clear from the question as input sequence is almost sorted. Now if you guess the peak point then you are in a good position, but during the binary search if you choose a number which will give you two cases.



        enter image description here



        The above two cases are the cases in which you need to think, but if you have tried few problems on binary search then now the problem will be easy to solve. More formally you can see the yuval's answer. You ask for the thought process, I have tried to give you my thought process. For any two persons thought process is different, there may be many other way to visualise the problem you have described.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          The first thing that comes to my mind after reading your question is as follow. Draw the sequence as a graph given below. Each $B[i]$ you can think of a point in the graph given below.
          enter image description here



          The idea to do binary search should be clear from the question as input sequence is almost sorted. Now if you guess the peak point then you are in a good position, but during the binary search if you choose a number which will give you two cases.



          enter image description here



          The above two cases are the cases in which you need to think, but if you have tried few problems on binary search then now the problem will be easy to solve. More formally you can see the yuval's answer. You ask for the thought process, I have tried to give you my thought process. For any two persons thought process is different, there may be many other way to visualise the problem you have described.






          share|cite|improve this answer









          $endgroup$



          The first thing that comes to my mind after reading your question is as follow. Draw the sequence as a graph given below. Each $B[i]$ you can think of a point in the graph given below.
          enter image description here



          The idea to do binary search should be clear from the question as input sequence is almost sorted. Now if you guess the peak point then you are in a good position, but during the binary search if you choose a number which will give you two cases.



          enter image description here



          The above two cases are the cases in which you need to think, but if you have tried few problems on binary search then now the problem will be easy to solve. More formally you can see the yuval's answer. You ask for the thought process, I have tried to give you my thought process. For any two persons thought process is different, there may be many other way to visualise the problem you have described.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 7 at 7:25









          aaagaaag

          1,142418




          1,142418























              1












              $begingroup$

              A sequence satisfying your constraint is known as bitonic. (Strictly speaking, a bitonic sequence is one which is either monotone increasing then decreasing or monotone decreasing then increasing). You can find the index $i^*$ using binary search. The idea is that given an index $i$, you can compare $i$ to $i^*$ as follows:




              • If $B[i-1] < B[i] > B[i+1]$, then $i = i^*$.

              • If $B[i-1] < B[i] < B[i+1]$, then $i < i^*$.

              • If $B[i-1] > B[i] > B[i+1]$, then $i > i^*$.


              I'll leave you to handle the corner cases $i = 1$ and $i = n$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                A sequence satisfying your constraint is known as bitonic. (Strictly speaking, a bitonic sequence is one which is either monotone increasing then decreasing or monotone decreasing then increasing). You can find the index $i^*$ using binary search. The idea is that given an index $i$, you can compare $i$ to $i^*$ as follows:




                • If $B[i-1] < B[i] > B[i+1]$, then $i = i^*$.

                • If $B[i-1] < B[i] < B[i+1]$, then $i < i^*$.

                • If $B[i-1] > B[i] > B[i+1]$, then $i > i^*$.


                I'll leave you to handle the corner cases $i = 1$ and $i = n$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  A sequence satisfying your constraint is known as bitonic. (Strictly speaking, a bitonic sequence is one which is either monotone increasing then decreasing or monotone decreasing then increasing). You can find the index $i^*$ using binary search. The idea is that given an index $i$, you can compare $i$ to $i^*$ as follows:




                  • If $B[i-1] < B[i] > B[i+1]$, then $i = i^*$.

                  • If $B[i-1] < B[i] < B[i+1]$, then $i < i^*$.

                  • If $B[i-1] > B[i] > B[i+1]$, then $i > i^*$.


                  I'll leave you to handle the corner cases $i = 1$ and $i = n$.






                  share|cite|improve this answer









                  $endgroup$



                  A sequence satisfying your constraint is known as bitonic. (Strictly speaking, a bitonic sequence is one which is either monotone increasing then decreasing or monotone decreasing then increasing). You can find the index $i^*$ using binary search. The idea is that given an index $i$, you can compare $i$ to $i^*$ as follows:




                  • If $B[i-1] < B[i] > B[i+1]$, then $i = i^*$.

                  • If $B[i-1] < B[i] < B[i+1]$, then $i < i^*$.

                  • If $B[i-1] > B[i] > B[i+1]$, then $i > i^*$.


                  I'll leave you to handle the corner cases $i = 1$ and $i = n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 7 at 4:49









                  Yuval FilmusYuval Filmus

                  195k14184349




                  195k14184349






























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