Ilgiausios bitoninės sekos spausdinimas

Ilgiausios bitoninės posekos problema yra rasti ilgiausią tam tikros sekos poseką, kad ji iš pradžių didėtų, o paskui mažėtų. Didėjančia tvarka surūšiuota seka laikoma bitine, o mažėjanti dalis tuščia. Panašiai mažėjančios eilės seka laikoma Bitonine, o didėjanti dalis tuščia. Pavyzdžiai:

  Input:    [1 11 2 10 4 5 2 1]   Output:   [1 2 10 4 2 1] OR [1 11 10 5 2 1] OR [1 2 4 5 2 1]   Input:    [12 11 40 5 3 1]   Output:   [12 11 5 3 1] OR [12 40 5 3 1]   Input:    [80 60 30 40 20 10]   Output:   [80 60 30 20 10] OR [80 60 40 20 10] 

Į ankstesnis pranešimas, kurį aptarėme apie ilgiausią bitoninę seką. Tačiau įrašas apėmė tik kodą, susijusį su didžiausios didėjančios eilės sumos radimu, bet ne su posekos konstravimu. Šiame įraše aptarsime, kaip sukurti pačią ilgiausią bitoninę seką. Tegul arr[0..n-1] yra įvesties masyvas. Vektorių LIS apibrėžiame taip, kad pats LIS[i] yra vektorius, kuris saugo ilgiausią didėjančią arr[0..i] seką, kuri baigiasi arr[i]. Todėl indeksui i LIS[i] gali būti rekursyviai parašytas kaip -

LIS[0] = {arr[O]} LIS[i] = {Max(LIS[j])} + arr[i] where   j  < i   and arr[j]  < arr[i] = arr[i] if there is no such j 

Taip pat apibrėžiame vektorių LDS, kad LDS[i] pats savaime būtų vektorius, kuris saugo ilgiausią mažėjančią arr[i..n] seką, kuri prasideda raide arr[i]. Todėl indeksui i LDS[i] gali būti rekursyviai parašytas kaip -

LDS[n] = {arr[n]} LDS[i] = arr[i] + {Max(LDS[j])} where   j > i   and arr[j]  < arr[i] = arr[i] if there is no such j 

Pavyzdžiui, masyvei [1 11 2 10 4 5 2 1]

LIS[0]: 1 LIS[1]: 1 11 LIS[2]: 1 2 LIS[3]: 1 2 10 LIS[4]: 1 2 4 LIS[5]: 1 2 4 5 LIS[6]: 1 2 LIS[7]: 1 
LDS[0]: 1 LDS[1]: 11 10 5 2 1 LDS[2]: 2 1 LDS[3]: 10 5 2 1 LDS[4]: 4 2 1 LDS[5]: 5 2 1 LDS[6]: 2 1 LDS[7]: 1 

Todėl gali būti ilgiausia bitoninė seka

LIS[1] + LDS[1] = [1 11 10 5 2 1] OR LIS[3] + LDS[3] = [1 2 10 5 2 1] OR LIS[5] + LDS[5] = [1 2 4 5 2 1] 

Žemiau yra aukščiau pateiktos idėjos įgyvendinimas - 

C++
   /* Dynamic Programming solution to print Longest    Bitonic Subsequence */   #include          using     namespace     std  ;   // Utility function to print Longest Bitonic   // Subsequence   void     print  (  vector   <  int  >&     arr       int     size  )   {      for  (  int     i     =     0  ;     i      <     size  ;     i  ++  )      cout      < <     arr  [  i  ]      < <     ' '  ;   }   // Function to construct and print Longest   // Bitonic Subsequence   void     printLBS  (  int     arr  []     int     n  )   {      // LIS[i] stores the length of the longest      // increasing subsequence ending with arr[i]      vector   <  vector   <  int  >>     LIS  (  n  );      // initialize LIS[0] to arr[0]      LIS  [  0  ].  push_back  (  arr  [  0  ]);      // Compute LIS values from left to right      for     (  int     i     =     1  ;     i      <     n  ;     i  ++  )      {      // for every j less than i      for     (  int     j     =     0  ;     j      <     i  ;     j  ++  )      {      if     ((  arr  [  j  ]      <     arr  [  i  ])     &&      (  LIS  [  j  ].  size  ()     >     LIS  [  i  ].  size  ()))      LIS  [  i  ]     =     LIS  [  j  ];      }      LIS  [  i  ].  push_back  (  arr  [  i  ]);      }      /* LIS[i] now stores Maximum Increasing    Subsequence of arr[0..i] that ends with    arr[i] */      // LDS[i] stores the length of the longest      // decreasing subsequence starting with arr[i]      vector   <  vector   <  int  >>     LDS  (  n  );      // initialize LDS[n-1] to arr[n-1]      LDS  [  n     -     1  ].  push_back  (  arr  [  n     -     1  ]);      // Compute LDS values from right to left      for     (  int     i     =     n     -     2  ;     i     >=     0  ;     i  --  )      {      // for every j greater than i      for     (  int     j     =     n     -     1  ;     j     >     i  ;     j  --  )      {      if     ((  arr  [  j  ]      <     arr  [  i  ])     &&      (  LDS  [  j  ].  size  ()     >     LDS  [  i  ].  size  ()))      LDS  [  i  ]     =     LDS  [  j  ];      }      LDS  [  i  ].  push_back  (  arr  [  i  ]);      }      // reverse as vector as we're inserting at end      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      reverse  (  LDS  [  i  ].  begin  ()     LDS  [  i  ].  end  ());      /* LDS[i] now stores Maximum Decreasing Subsequence    of arr[i..n] that starts with arr[i] */      int     max     =     0  ;      int     maxIndex     =     -1  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      {      // Find maximum value of size of LIS[i] + size      // of LDS[i] - 1      if     (  LIS  [  i  ].  size  ()     +     LDS  [  i  ].  size  ()     -     1     >     max  )      {      max     =     LIS  [  i  ].  size  ()     +     LDS  [  i  ].  size  ()     -     1  ;      maxIndex     =     i  ;      }      }      // print all but last element of LIS[maxIndex] vector      print  (  LIS  [  maxIndex  ]     LIS  [  maxIndex  ].  size  ()     -     1  );      // print all elements of LDS[maxIndex] vector      print  (  LDS  [  maxIndex  ]     LDS  [  maxIndex  ].  size  ());   }   // Driver program   int     main  ()   {      int     arr  []     =     {     1       11       2       10       4       5       2       1     };      int     n     =     sizeof  (  arr  )     /     sizeof  (  arr  [  0  ]);      printLBS  (  arr       n  );      return     0  ;   }   
Java
   /* Dynamic Programming solution to print Longest    Bitonic Subsequence */   import     java.util.*  ;   class   GFG      {      // Utility function to print Longest Bitonic      // Subsequence      static     void     print  (  Vector   <  Integer  >     arr       int     size  )         {      for     (  int     i     =     0  ;     i      <     size  ;     i  ++  )      System  .  out  .  print  (  arr  .  elementAt  (  i  )     +     ' '  );      }      // Function to construct and print Longest      // Bitonic Subsequence      static     void     printLBS  (  int  []     arr       int     n  )         {      // LIS[i] stores the length of the longest      // increasing subsequence ending with arr[i]      @SuppressWarnings  (  'unchecked'  )      Vector   <  Integer  >[]     LIS     =     new     Vector  [  n  ]  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      LIS  [  i  ]     =     new     Vector   <>  ();      // initialize LIS[0] to arr[0]      LIS  [  0  ]  .  add  (  arr  [  0  ]  );      // Compute LIS values from left to right      for     (  int     i     =     1  ;     i      <     n  ;     i  ++  )         {      // for every j less than i      for     (  int     j     =     0  ;     j      <     i  ;     j  ++  )         {      if     ((  arr  [  i  ]     >     arr  [  j  ]  )     &&         LIS  [  j  ]  .  size  ()     >     LIS  [  i  ]  .  size  ())         {      for     (  int     k     :     LIS  [  j  ]  )      if     (  !  LIS  [  i  ]  .  contains  (  k  ))      LIS  [  i  ]  .  add  (  k  );      }      }      LIS  [  i  ]  .  add  (  arr  [  i  ]  );      }      /*    * LIS[i] now stores Maximum Increasing Subsequence     * of arr[0..i] that ends with arr[i]    */      // LDS[i] stores the length of the longest      // decreasing subsequence starting with arr[i]      @SuppressWarnings  (  'unchecked'  )      Vector   <  Integer  >[]     LDS     =     new     Vector  [  n  ]  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      LDS  [  i  ]     =     new     Vector   <>  ();      // initialize LDS[n-1] to arr[n-1]      LDS  [  n     -     1  ]  .  add  (  arr  [  n     -     1  ]  );      // Compute LDS values from right to left      for     (  int     i     =     n     -     2  ;     i     >=     0  ;     i  --  )         {      // for every j greater than i      for     (  int     j     =     n     -     1  ;     j     >     i  ;     j  --  )         {      if     (  arr  [  j  ]      <     arr  [  i  ]     &&         LDS  [  j  ]  .  size  ()     >     LDS  [  i  ]  .  size  ())      for     (  int     k     :     LDS  [  j  ]  )      if     (  !  LDS  [  i  ]  .  contains  (  k  ))      LDS  [  i  ]  .  add  (  k  );      }      LDS  [  i  ]  .  add  (  arr  [  i  ]  );      }      // reverse as vector as we're inserting at end      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      Collections  .  reverse  (  LDS  [  i  ]  );      /*    * LDS[i] now stores Maximum Decreasing Subsequence     * of arr[i..n] that starts with arr[i]    */      int     max     =     0  ;      int     maxIndex     =     -  1  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      {      // Find maximum value of size of       // LIS[i] + size of LDS[i] - 1      if     (  LIS  [  i  ]  .  size  ()     +     LDS  [  i  ]  .  size  ()     -     1     >     max  )      {      max     =     LIS  [  i  ]  .  size  ()     +     LDS  [  i  ]  .  size  ()     -     1  ;      maxIndex     =     i  ;      }      }      // print all but last element of LIS[maxIndex] vector      print  (  LIS  [  maxIndex  ]       LIS  [  maxIndex  ]  .  size  ()     -     1  );      // print all elements of LDS[maxIndex] vector      print  (  LDS  [  maxIndex  ]       LDS  [  maxIndex  ]  .  size  ());      }      // Driver Code      public     static     void     main  (  String  []     args  )         {      int  []     arr     =     {     1       11       2       10       4       5       2       1     };      int     n     =     arr  .  length  ;      printLBS  (  arr       n  );      }   }   // This code is contributed by   // sanjeev2552   
Python3
   # Dynamic Programming solution to print Longest   # Bitonic Subsequence   def   _print  (  arr  :   list     size  :   int  ):   for   i   in   range  (  size  ):   print  (  arr  [  i  ]   end  =  ' '  )   # Function to construct and print Longest   # Bitonic Subsequence   def   printLBS  (  arr  :   list     n  :   int  ):   # LIS[i] stores the length of the longest   # increasing subsequence ending with arr[i]   LIS   =   [  0  ]   *   n   for   i   in   range  (  n  ):   LIS  [  i  ]   =   []   # initialize LIS[0] to arr[0]   LIS  [  0  ]  .  append  (  arr  [  0  ])   # Compute LIS values from left to right   for   i   in   range  (  1     n  ):   # for every j less than i   for   j   in   range  (  i  ):   if   ((  arr  [  j  ]    <   arr  [  i  ])   and   (  len  (  LIS  [  j  ])   >   len  (  LIS  [  i  ]))):   LIS  [  i  ]   =   LIS  [  j  ]  .  copy  ()   LIS  [  i  ]  .  append  (  arr  [  i  ])   # LIS[i] now stores Maximum Increasing   # Subsequence of arr[0..i] that ends with   # arr[i]   # LDS[i] stores the length of the longest   # decreasing subsequence starting with arr[i]   LDS   =   [  0  ]   *   n   for   i   in   range  (  n  ):   LDS  [  i  ]   =   []   # initialize LDS[n-1] to arr[n-1]   LDS  [  n   -   1  ]  .  append  (  arr  [  n   -   1  ])   # Compute LDS values from right to left   for   i   in   range  (  n   -   2     -  1     -  1  ):   # for every j greater than i   for   j   in   range  (  n   -   1     i     -  1  ):   if   ((  arr  [  j  ]    <   arr  [  i  ])   and   (  len  (  LDS  [  j  ])   >   len  (  LDS  [  i  ]))):   LDS  [  i  ]   =   LDS  [  j  ]  .  copy  ()   LDS  [  i  ]  .  append  (  arr  [  i  ])   # reverse as vector as we're inserting at end   for   i   in   range  (  n  ):   LDS  [  i  ]   =   list  (  reversed  (  LDS  [  i  ]))   # LDS[i] now stores Maximum Decreasing Subsequence   # of arr[i..n] that starts with arr[i]   max   =   0   maxIndex   =   -  1   for   i   in   range  (  n  ):   # Find maximum value of size of LIS[i] + size   # of LDS[i] - 1   if   (  len  (  LIS  [  i  ])   +   len  (  LDS  [  i  ])   -   1   >   max  ):   max   =   len  (  LIS  [  i  ])   +   len  (  LDS  [  i  ])   -   1   maxIndex   =   i   # print all but last element of LIS[maxIndex] vector   _print  (  LIS  [  maxIndex  ]   len  (  LIS  [  maxIndex  ])   -   1  )   # print all elements of LDS[maxIndex] vector   _print  (  LDS  [  maxIndex  ]   len  (  LDS  [  maxIndex  ]))   # Driver Code   if   __name__   ==   '__main__'  :   arr   =   [  1     11     2     10     4     5     2     1  ]   n   =   len  (  arr  )   printLBS  (  arr     n  )   # This code is contributed by   # sanjeev2552   
C#
   /* Dynamic Programming solution to print longest    Bitonic Subsequence */   using     System  ;   using     System.Linq  ;   using     System.Collections.Generic  ;   class     GFG      {      // Utility function to print longest Bitonic      // Subsequence      static     void     print  (  List   <  int  >     arr       int     size  )         {      for     (  int     i     =     0  ;     i      <     size  ;     i  ++  )      Console  .  Write  (  arr  [  i  ]     +     ' '  );      }      // Function to construct and print longest      // Bitonic Subsequence      static     void     printLBS  (  int  []     arr       int     n  )         {      // LIS[i] stores the length of the longest      // increasing subsequence ending with arr[i]      List   <  int  >  []     LIS     =     new     List   <  int  >  [  n  ];      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      LIS  [  i  ]     =     new     List   <  int  >  ();      // initialize LIS[0] to arr[0]      LIS  [  0  ].  Add  (  arr  [  0  ]);      // Compute LIS values from left to right      for     (  int     i     =     1  ;     i      <     n  ;     i  ++  )         {      // for every j less than i      for     (  int     j     =     0  ;     j      <     i  ;     j  ++  )         {      if     ((  arr  [  i  ]     >     arr  [  j  ])     &&         LIS  [  j  ].  Count     >     LIS  [  i  ].  Count  )         {      foreach     (  int     k     in     LIS  [  j  ])      if     (  !  LIS  [  i  ].  Contains  (  k  ))      LIS  [  i  ].  Add  (  k  );      }      }      LIS  [  i  ].  Add  (  arr  [  i  ]);      }      /*    * LIS[i] now stores Maximum Increasing Subsequence     * of arr[0..i] that ends with arr[i]    */      // LDS[i] stores the length of the longest      // decreasing subsequence starting with arr[i]      List   <  int  >  []     LDS     =     new     List   <  int  >  [  n  ];      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      LDS  [  i  ]     =     new     List   <  int  >  ();      // initialize LDS[n-1] to arr[n-1]      LDS  [  n     -     1  ].  Add  (  arr  [  n     -     1  ]);      // Compute LDS values from right to left      for     (  int     i     =     n     -     2  ;     i     >=     0  ;     i  --  )         {      // for every j greater than i      for     (  int     j     =     n     -     1  ;     j     >     i  ;     j  --  )         {      if     (  arr  [  j  ]      <     arr  [  i  ]     &&         LDS  [  j  ].  Count     >     LDS  [  i  ].  Count  )      foreach     (  int     k     in     LDS  [  j  ])      if     (  !  LDS  [  i  ].  Contains  (  k  ))      LDS  [  i  ].  Add  (  k  );      }      LDS  [  i  ].  Add  (  arr  [  i  ]);      }      // reverse as vector as we're inserting at end      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      LDS  [  i  ].  Reverse  ();      /*    * LDS[i] now stores Maximum Decreasing Subsequence     * of arr[i..n] that starts with arr[i]    */      int     max     =     0  ;         int     maxIndex     =     -  1  ;      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      {      // Find maximum value of size of       // LIS[i] + size of LDS[i] - 1      if     (  LIS  [  i  ].  Count     +     LDS  [  i  ].  Count     -     1     >     max  )      {      max     =     LIS  [  i  ].  Count     +     LDS  [  i  ].  Count     -     1  ;      maxIndex     =     i  ;      }      }      // print all but last element of LIS[maxIndex] vector      print  (  LIS  [  maxIndex  ]     LIS  [  maxIndex  ].  Count     -     1  );      // print all elements of LDS[maxIndex] vector      print  (  LDS  [  maxIndex  ]     LDS  [  maxIndex  ].  Count  );      }      // Driver Code      public     static     void     Main  (  String  []     args  )         {      int  []     arr     =     {     1       11       2       10       4       5       2       1     };      int     n     =     arr  .  Length  ;      printLBS  (  arr       n  );      }   }   // This code is contributed by PrinciRaj1992   
JavaScript
   // Function to print the longest bitonic subsequence   function     _print  (  arr       size  )     {      for     (  let     i     =     0  ;     i   <  size  ;     i  ++  )     {      process  .  stdout  .  write  (  arr  [  i  ]  +  ' '  );      }   }   // Function to construct and print the longest bitonic subsequence   function     printLBS  (  arr       n  )     {      // LIS[i] stores the length of the longest increasing subsequence ending with arr[i]      let     LIS     =     new     Array  (  n  );      for     (  let     i     =     0  ;     i      <     n  ;     i  ++  )     {      LIS  [  i  ]     =     [];      }      // initialize LIS[0] to arr[0]      LIS  [  0  ].  push  (  arr  [  0  ]);      // Compute LIS values from left to right      for     (  let     i     =     1  ;     i      <     n  ;     i  ++  )     {      // for every j less than i      for     (  let     j     =     0  ;     j      <     i  ;     j  ++  )     {      if     (  arr  [  j  ]      <     arr  [  i  ]     &&     LIS  [  j  ].  length     >     LIS  [  i  ].  length  )     {      LIS  [  i  ]     =     LIS  [  j  ].  slice  ();      }      }      LIS  [  i  ].  push  (  arr  [  i  ]);      }      // LIS[i] now stores the Maximum Increasing Subsequence of arr[0..i] that ends with arr[i]      // LDS[i] stores the length of the longest decreasing subsequence starting with arr[i]      let     LDS     =     new     Array  (  n  );      for     (  let     i     =     0  ;     i      <     n  ;     i  ++  )     {      LDS  [  i  ]     =     [];      }      // initialize LDS[n-1] to arr[n-1]      LDS  [  n     -     1  ].  push  (  arr  [  n     -     1  ]);      // Compute LDS values from right to left      for     (  let     i     =     n     -     2  ;     i     >=     0  ;     i  --  )     {      // for every j greater than i      for     (  let     j     =     n     -     1  ;     j     >     i  ;     j  --  )     {      if     (  arr  [  j  ]      <     arr  [  i  ]     &&     LDS  [  j  ].  length     >     LDS  [  i  ].  length  )     {      LDS  [  i  ]     =     LDS  [  j  ].  slice  ();      }      }      LDS  [  i  ].  push  (  arr  [  i  ]);      }      // reverse the LDS vector as we're inserting at the end      for     (  let     i     =     0  ;     i      <     n  ;     i  ++  )     {      LDS  [  i  ].  reverse  ();      }      // LDS[i] now stores the Maximum Decreasing Subsequence of arr[i..n] that starts with arr[i]      let     max     =     0  ;      let     maxIndex     =     -  1  ;      for     (  let     i     =     0  ;     i      <     n  ;     i  ++  )     {      // Find maximum value of size of LIS[i] + size of LDS[i] - 1      if     (  LIS  [  i  ].  length     +     LDS  [  i  ].  length     -     1     >     max  )     {      max     =     LIS  [  i  ].  length     +     LDS  [  i  ].  length     -     1  ;      maxIndex     =     i  ;      }      }      // print all but      // print all but last element of LIS[maxIndex] array      _print  (  LIS  [  maxIndex  ].  slice  (  0       -  1  )     LIS  [  maxIndex  ].  length     -     1  );      // print all elements of LDS[maxIndex] array      _print  (  LDS  [  maxIndex  ]     LDS  [  maxIndex  ].  length  );   }   // Driver program   const     arr     =     [  1       11       2       10       4       5       2       1  ];   const     n     =     arr  .  length  ;   printLBS  (  arr       n  );   

Išvestis:

1 11 10 5 2 1 

Laiko sudėtingumas dinaminio programavimo sprendimas yra O(n 2 ). Pagalbinė erdvė programos naudojamas O (n 2 ).