Index matching algorithm without hash-based data structures?












6












$begingroup$


I am programming in C, so I do not want to implement a hash-based datastructure such as a hashset or hashmap/dictionary. However, I need to solve the following task in linear time.



Given two arrays $a$ and $b$ which contain the same set of distinct integers, determine for every element of $a$ the index of the same element in $b$.



For example, if $a=[9,4,3,7]$ and $b=[3,4,7,9]$, then the output should be $[3,1,0,2]$.



Note that this becomes a very easy task when you have a hashset, because you can simply store for every element in $b$ the index, and then query the hashmap for every element of $a$.



So my question is whether there is a linear algorithm for this task that does not use any hashsets.






search-algorithms hash-tables permutations







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    6












    $begingroup$


    I am programming in C, so I do not want to implement a hash-based datastructure such as a hashset or hashmap/dictionary. However, I need to solve the following task in linear time.



    Given two arrays $a$ and $b$ which contain the same set of distinct integers, determine for every element of $a$ the index of the same element in $b$.



    For example, if $a=[9,4,3,7]$ and $b=[3,4,7,9]$, then the output should be $[3,1,0,2]$.



    Note that this becomes a very easy task when you have a hashset, because you can simply store for every element in $b$ the index, and then query the hashmap for every element of $a$.



    So my question is whether there is a linear algorithm for this task that does not use any hashsets.






    search-algorithms hash-tables permutations







    share|cite|improve this question







    New contributor




    SmileyCraft is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      6












      6








      6





      $begingroup$


      I am programming in C, so I do not want to implement a hash-based datastructure such as a hashset or hashmap/dictionary. However, I need to solve the following task in linear time.



      Given two arrays $a$ and $b$ which contain the same set of distinct integers, determine for every element of $a$ the index of the same element in $b$.



      For example, if $a=[9,4,3,7]$ and $b=[3,4,7,9]$, then the output should be $[3,1,0,2]$.



      Note that this becomes a very easy task when you have a hashset, because you can simply store for every element in $b$ the index, and then query the hashmap for every element of $a$.



      So my question is whether there is a linear algorithm for this task that does not use any hashsets.






      search-algorithms hash-tables permutations







      share|cite|improve this question







      New contributor




      SmileyCraft is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I am programming in C, so I do not want to implement a hash-based datastructure such as a hashset or hashmap/dictionary. However, I need to solve the following task in linear time.



      Given two arrays $a$ and $b$ which contain the same set of distinct integers, determine for every element of $a$ the index of the same element in $b$.



      For example, if $a=[9,4,3,7]$ and $b=[3,4,7,9]$, then the output should be $[3,1,0,2]$.



      Note that this becomes a very easy task when you have a hashset, because you can simply store for every element in $b$ the index, and then query the hashmap for every element of $a$.



      So my question is whether there is a linear algorithm for this task that does not use any hashsets.






      search-algorithms hash-tables permutations




      search-algorithms hash-tables permutations






      share|cite|improve this question







      New contributor




      SmileyCraft is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      SmileyCraft is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






      New contributor




      SmileyCraft is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 6 hours ago









      SmileyCraftSmileyCraft

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      New contributor




      SmileyCraft is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      SmileyCraft is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.






















          1 Answer
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          active

          oldest

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          2












          $begingroup$

          If the only operation allowed between any two (possibly the same) elements in the two arrays is to determine which one is the smaller one, then it will take $Theta(nlog n)$ time in worst case for any algorithm. This can be seen in the case when array $a$ is sorted while array $b$ is not. Then knowing the index $I(k)$ of the same element in $b$ for the $k$-th element of $a$ for all $k$, we can sort $b$ in linear time by putting $b_{I(k)}$ in $k$-th position.



          The following is a formal formulation of the conclusion above in the comparison computation model.



          Let $mathcal O$ be an oracle that can tell a fixed strict linear ordering on $E$, a set of $n$ elements. That is, on input $e,fin E$, $mathcal O$ outputs -1 if $eprec f$, 0 if $e$ is $f$ and 1 otherwise. Let $a$ and $b$ are two bijections from ${0, 1,cdots, n-1}$ to $E$. To output $I(0), I(1), cdots, I(n-1)$ in that order such that $a(k)=b(I(k))$ for all $0le kle n-1$, it will take $Theta(nlog n)$ queries against $mathcal O$ in the worst case.




          whether there is a linear algorithm for this task that does not use any hashsets.




          A computation model that is defined by no usage of hashset is not a well-defined computation mode. How can you check there is no usage of hashset? There are literally hundreds of ways to implement a data structure that is a hashset or looks like a hashset or looks like a hashset partially. In general, a well-defined computation model must be defined by what can be done formally.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I don't think OP is asking for an ordering of the elements of a. It sounds more like he is asking for a mapping i.e. map element of a to it's position in b; not order elements of a according to their position in b. Ordering will require O(nlogn) as you have astutely observed, but mapping can be done in O(n).
            $endgroup$
            – smac89
            2 hours ago










          • $begingroup$
            Exactly, I don't think OP is asking for an ordering of the element of $a$. Please read my answer carefully, especially the formal formulation. Please come to chat.stackexchange.com/rooms/2710/computer-science for a chat.
            $endgroup$
            – Apass.Jack
            2 hours ago










          • $begingroup$
            This answer abstracts "distinct integers" as "distinct elements" with a strict total order. There could be other computation models for "distinct integers" without "hashset".
            $endgroup$
            – Apass.Jack
            1 hour ago











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          active

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          2












          $begingroup$

          If the only operation allowed between any two (possibly the same) elements in the two arrays is to determine which one is the smaller one, then it will take $Theta(nlog n)$ time in worst case for any algorithm. This can be seen in the case when array $a$ is sorted while array $b$ is not. Then knowing the index $I(k)$ of the same element in $b$ for the $k$-th element of $a$ for all $k$, we can sort $b$ in linear time by putting $b_{I(k)}$ in $k$-th position.



          The following is a formal formulation of the conclusion above in the comparison computation model.



          Let $mathcal O$ be an oracle that can tell a fixed strict linear ordering on $E$, a set of $n$ elements. That is, on input $e,fin E$, $mathcal O$ outputs -1 if $eprec f$, 0 if $e$ is $f$ and 1 otherwise. Let $a$ and $b$ are two bijections from ${0, 1,cdots, n-1}$ to $E$. To output $I(0), I(1), cdots, I(n-1)$ in that order such that $a(k)=b(I(k))$ for all $0le kle n-1$, it will take $Theta(nlog n)$ queries against $mathcal O$ in the worst case.




          whether there is a linear algorithm for this task that does not use any hashsets.




          A computation model that is defined by no usage of hashset is not a well-defined computation mode. How can you check there is no usage of hashset? There are literally hundreds of ways to implement a data structure that is a hashset or looks like a hashset or looks like a hashset partially. In general, a well-defined computation model must be defined by what can be done formally.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I don't think OP is asking for an ordering of the elements of a. It sounds more like he is asking for a mapping i.e. map element of a to it's position in b; not order elements of a according to their position in b. Ordering will require O(nlogn) as you have astutely observed, but mapping can be done in O(n).
            $endgroup$
            – smac89
            2 hours ago










          • $begingroup$
            Exactly, I don't think OP is asking for an ordering of the element of $a$. Please read my answer carefully, especially the formal formulation. Please come to chat.stackexchange.com/rooms/2710/computer-science for a chat.
            $endgroup$
            – Apass.Jack
            2 hours ago










          • $begingroup$
            This answer abstracts "distinct integers" as "distinct elements" with a strict total order. There could be other computation models for "distinct integers" without "hashset".
            $endgroup$
            – Apass.Jack
            1 hour ago
















          2












          $begingroup$

          If the only operation allowed between any two (possibly the same) elements in the two arrays is to determine which one is the smaller one, then it will take $Theta(nlog n)$ time in worst case for any algorithm. This can be seen in the case when array $a$ is sorted while array $b$ is not. Then knowing the index $I(k)$ of the same element in $b$ for the $k$-th element of $a$ for all $k$, we can sort $b$ in linear time by putting $b_{I(k)}$ in $k$-th position.



          The following is a formal formulation of the conclusion above in the comparison computation model.



          Let $mathcal O$ be an oracle that can tell a fixed strict linear ordering on $E$, a set of $n$ elements. That is, on input $e,fin E$, $mathcal O$ outputs -1 if $eprec f$, 0 if $e$ is $f$ and 1 otherwise. Let $a$ and $b$ are two bijections from ${0, 1,cdots, n-1}$ to $E$. To output $I(0), I(1), cdots, I(n-1)$ in that order such that $a(k)=b(I(k))$ for all $0le kle n-1$, it will take $Theta(nlog n)$ queries against $mathcal O$ in the worst case.




          whether there is a linear algorithm for this task that does not use any hashsets.




          A computation model that is defined by no usage of hashset is not a well-defined computation mode. How can you check there is no usage of hashset? There are literally hundreds of ways to implement a data structure that is a hashset or looks like a hashset or looks like a hashset partially. In general, a well-defined computation model must be defined by what can be done formally.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I don't think OP is asking for an ordering of the elements of a. It sounds more like he is asking for a mapping i.e. map element of a to it's position in b; not order elements of a according to their position in b. Ordering will require O(nlogn) as you have astutely observed, but mapping can be done in O(n).
            $endgroup$
            – smac89
            2 hours ago










          • $begingroup$
            Exactly, I don't think OP is asking for an ordering of the element of $a$. Please read my answer carefully, especially the formal formulation. Please come to chat.stackexchange.com/rooms/2710/computer-science for a chat.
            $endgroup$
            – Apass.Jack
            2 hours ago










          • $begingroup$
            This answer abstracts "distinct integers" as "distinct elements" with a strict total order. There could be other computation models for "distinct integers" without "hashset".
            $endgroup$
            – Apass.Jack
            1 hour ago














          2












          2








          2





          $begingroup$

          If the only operation allowed between any two (possibly the same) elements in the two arrays is to determine which one is the smaller one, then it will take $Theta(nlog n)$ time in worst case for any algorithm. This can be seen in the case when array $a$ is sorted while array $b$ is not. Then knowing the index $I(k)$ of the same element in $b$ for the $k$-th element of $a$ for all $k$, we can sort $b$ in linear time by putting $b_{I(k)}$ in $k$-th position.



          The following is a formal formulation of the conclusion above in the comparison computation model.



          Let $mathcal O$ be an oracle that can tell a fixed strict linear ordering on $E$, a set of $n$ elements. That is, on input $e,fin E$, $mathcal O$ outputs -1 if $eprec f$, 0 if $e$ is $f$ and 1 otherwise. Let $a$ and $b$ are two bijections from ${0, 1,cdots, n-1}$ to $E$. To output $I(0), I(1), cdots, I(n-1)$ in that order such that $a(k)=b(I(k))$ for all $0le kle n-1$, it will take $Theta(nlog n)$ queries against $mathcal O$ in the worst case.




          whether there is a linear algorithm for this task that does not use any hashsets.




          A computation model that is defined by no usage of hashset is not a well-defined computation mode. How can you check there is no usage of hashset? There are literally hundreds of ways to implement a data structure that is a hashset or looks like a hashset or looks like a hashset partially. In general, a well-defined computation model must be defined by what can be done formally.






          share|cite|improve this answer











          $endgroup$



          If the only operation allowed between any two (possibly the same) elements in the two arrays is to determine which one is the smaller one, then it will take $Theta(nlog n)$ time in worst case for any algorithm. This can be seen in the case when array $a$ is sorted while array $b$ is not. Then knowing the index $I(k)$ of the same element in $b$ for the $k$-th element of $a$ for all $k$, we can sort $b$ in linear time by putting $b_{I(k)}$ in $k$-th position.



          The following is a formal formulation of the conclusion above in the comparison computation model.



          Let $mathcal O$ be an oracle that can tell a fixed strict linear ordering on $E$, a set of $n$ elements. That is, on input $e,fin E$, $mathcal O$ outputs -1 if $eprec f$, 0 if $e$ is $f$ and 1 otherwise. Let $a$ and $b$ are two bijections from ${0, 1,cdots, n-1}$ to $E$. To output $I(0), I(1), cdots, I(n-1)$ in that order such that $a(k)=b(I(k))$ for all $0le kle n-1$, it will take $Theta(nlog n)$ queries against $mathcal O$ in the worst case.




          whether there is a linear algorithm for this task that does not use any hashsets.




          A computation model that is defined by no usage of hashset is not a well-defined computation mode. How can you check there is no usage of hashset? There are literally hundreds of ways to implement a data structure that is a hashset or looks like a hashset or looks like a hashset partially. In general, a well-defined computation model must be defined by what can be done formally.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 2 hours ago









          Apass.JackApass.Jack

          13k1939




          13k1939












          • $begingroup$
            I don't think OP is asking for an ordering of the elements of a. It sounds more like he is asking for a mapping i.e. map element of a to it's position in b; not order elements of a according to their position in b. Ordering will require O(nlogn) as you have astutely observed, but mapping can be done in O(n).
            $endgroup$
            – smac89
            2 hours ago










          • $begingroup$
            Exactly, I don't think OP is asking for an ordering of the element of $a$. Please read my answer carefully, especially the formal formulation. Please come to chat.stackexchange.com/rooms/2710/computer-science for a chat.
            $endgroup$
            – Apass.Jack
            2 hours ago










          • $begingroup$
            This answer abstracts "distinct integers" as "distinct elements" with a strict total order. There could be other computation models for "distinct integers" without "hashset".
            $endgroup$
            – Apass.Jack
            1 hour ago


















          • $begingroup$
            I don't think OP is asking for an ordering of the elements of a. It sounds more like he is asking for a mapping i.e. map element of a to it's position in b; not order elements of a according to their position in b. Ordering will require O(nlogn) as you have astutely observed, but mapping can be done in O(n).
            $endgroup$
            – smac89
            2 hours ago










          • $begingroup$
            Exactly, I don't think OP is asking for an ordering of the element of $a$. Please read my answer carefully, especially the formal formulation. Please come to chat.stackexchange.com/rooms/2710/computer-science for a chat.
            $endgroup$
            – Apass.Jack
            2 hours ago










          • $begingroup$
            This answer abstracts "distinct integers" as "distinct elements" with a strict total order. There could be other computation models for "distinct integers" without "hashset".
            $endgroup$
            – Apass.Jack
            1 hour ago
















          $begingroup$
          I don't think OP is asking for an ordering of the elements of a. It sounds more like he is asking for a mapping i.e. map element of a to it's position in b; not order elements of a according to their position in b. Ordering will require O(nlogn) as you have astutely observed, but mapping can be done in O(n).
          $endgroup$
          – smac89
          2 hours ago




          $begingroup$
          I don't think OP is asking for an ordering of the elements of a. It sounds more like he is asking for a mapping i.e. map element of a to it's position in b; not order elements of a according to their position in b. Ordering will require O(nlogn) as you have astutely observed, but mapping can be done in O(n).
          $endgroup$
          – smac89
          2 hours ago












          $begingroup$
          Exactly, I don't think OP is asking for an ordering of the element of $a$. Please read my answer carefully, especially the formal formulation. Please come to chat.stackexchange.com/rooms/2710/computer-science for a chat.
          $endgroup$
          – Apass.Jack
          2 hours ago




          $begingroup$
          Exactly, I don't think OP is asking for an ordering of the element of $a$. Please read my answer carefully, especially the formal formulation. Please come to chat.stackexchange.com/rooms/2710/computer-science for a chat.
          $endgroup$
          – Apass.Jack
          2 hours ago












          $begingroup$
          This answer abstracts "distinct integers" as "distinct elements" with a strict total order. There could be other computation models for "distinct integers" without "hashset".
          $endgroup$
          – Apass.Jack
          1 hour ago




          $begingroup$
          This answer abstracts "distinct integers" as "distinct elements" with a strict total order. There could be other computation models for "distinct integers" without "hashset".
          $endgroup$
          – Apass.Jack
          1 hour ago










          SmileyCraft is a new contributor. Be nice, and check out our Code of Conduct.










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