Imprimiu tots els nombres d'n dígits amb la diferència de suma de dígits parells i senars com a 1

Donat un nombre enter n que representa el nombre de dígits. La tasca és imprimir-ho tot nombres de n xifres de manera que la diferència absoluta entre la suma de dígits a les posicions parells i les posicions senars és exactament 1 .
Nota : El número no ha de començar per (no es permeten zeros inicials).

Exemples:  

Entrada : n = 2
Sortida : 10 12 21 23 32 34 43 45 54 56 65 67 76 78 87 89 98

Entrada : n = 3
Sortida : 100 111 120 122 131 133 142 144 153 155 164 166 175 177 186
188 197 199 210 221 230 232 241 243 252 254 263 265 274 276 285
287 296 298 320 331 340 342 351 353 362 364 373 375 384 386 395
397 430 441 450 452 461 463 472 474 483 485 494 496 540 551 560
562 571 573 582 584 593 595 650 661 670 672 681 683 692 694 760
771 780 782 791 793 870 881 890 892 980 991  

[Enfocament esperat] Ús de la recursió

La idea és fer recursivament genera tots els nombres de n dígits mentre seguiment de la suma de dígits a fins i tot i estrany posicions utilitzant dues variables. Per a una posició determinada, l'omplim amb tots els dígits del 0 al 9 i en funció de si la posició actual és parell o senar, incrementem la suma parell o senar. Tractem els 0 inicials per separat, ja que no es compten com a dígits.
Hem seguit la numeració basada en zero com els índexs de matriu. és a dir, el dígit inicial (l'extrem esquerre) es considera present a la posició parell i el dígit al costat es considera en la posició senar, etc.

C++
   // C++ program to print all n-digit numbers such that   // the absolute difference between the sum of digits at   // even and odd positions is 1   #include          using     namespace     std  ;   // Recursive function to generate numbers   void     findNDigitNumsUtil  (  int     pos       int     n       int     num        int     evenSum       int     oddSum        vector   <  int  >     &  res  )     {      // If number is formed      if     (  pos     ==     n  )     {      // Check absolute difference condition      if     (  abs  (  evenSum     -     oddSum  )     ==     1  )     {      res  .  push_back  (  num  );      }      return  ;      }      // Digits to consider at current position      for     (  int     d     =     0  ;     d      <=     9  ;     d  ++  )     {      // Skip leading 0      if     (  pos     ==     0     &&     d     ==     0  )     {      continue  ;      }      // If position is even (0-based) add to evenSum      if     (  pos     %     2     ==     0  )     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum     +     d       oddSum       res  );      }      // If position is odd add to oddSum      else     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum       oddSum     +     d       res  );      }      }   }   // Function to prepare and collect valid numbers   vector   <  int  >     findNDigitNums  (  int     n  )     {          vector   <  int  >     res  ;      findNDigitNumsUtil  (  0       n       0       0       0       res  );          return     res  ;   }   // Driver code   int     main  ()     {      int     n     =     2  ;      vector   <  int  >     res     =     findNDigitNums  (  n  );      for     (  int     i     =     0  ;     i      <     res  .  size  ();     i  ++  )     {      cout      < <     res  [  i  ]      < <     ' '  ;      }      return     0  ;   }   
Java
   // Java program to print all n-digit numbers such that   // the absolute difference between the sum of digits at   // even and odd positions is 1   import     java.util.*  ;   class   GfG     {      // Recursive function to generate numbers      static     void     findNDigitNumsUtil  (  int     pos       int     n       int     num        int     evenSum       int     oddSum        ArrayList   <  Integer  >     res  )     {      // If number is formed      if     (  pos     ==     n  )     {      // Check absolute difference condition      if     (  Math  .  abs  (  evenSum     -     oddSum  )     ==     1  )     {      res  .  add  (  num  );      }      return  ;      }      // Digits to consider at current position      for     (  int     d     =     0  ;     d      <=     9  ;     d  ++  )     {      // Skip leading 0      if     (  pos     ==     0     &&     d     ==     0  )     {      continue  ;      }      // If position is even (0-based) add to evenSum      if     (  pos     %     2     ==     0  )     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum     +     d       oddSum       res  );      }      // If position is odd add to oddSum      else     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum       oddSum     +     d       res  );      }      }      }      // Function to prepare and collect valid numbers      static     ArrayList   <  Integer  >     findNDigitNums  (  int     n  )     {      ArrayList   <  Integer  >     res     =     new     ArrayList   <>  ();      findNDigitNumsUtil  (  0       n       0       0       0       res  );      return     res  ;      }      // Driver code      public     static     void     main  (  String  []     args  )     {      int     n     =     2  ;      ArrayList   <  Integer  >     res     =     findNDigitNums  (  n  );      // Print all collected valid numbers      for     (  int     i     =     0  ;     i      <     res  .  size  ();     i  ++  )     {      System  .  out  .  print  (  res  .  get  (  i  )     +     ' '  );      }      }   }   
Python
   # Python program to print all n-digit numbers such that   # the absolute difference between the sum of digits at   # even and odd positions is 1   # Recursive function to generate numbers   def   findNDigitNumsUtil  (  pos     n     num     evenSum     oddSum     res  ):   # If number is formed   if   pos   ==   n  :   # Check absolute difference condition   if   abs  (  evenSum   -   oddSum  )   ==   1  :   res  .  append  (  num  )   return   # Digits to consider at current position   for   d   in   range  (  10  ):   # Skip leading 0   if   pos   ==   0   and   d   ==   0  :   continue   # If position is even (0-based) add to evenSum   if   pos   %   2   ==   0  :   findNDigitNumsUtil  (  pos   +   1     n     num   *   10   +   d     evenSum   +   d     oddSum     res  )   # If position is odd add to oddSum   else  :   findNDigitNumsUtil  (  pos   +   1     n     num   *   10   +   d     evenSum     oddSum   +   d     res  )   # Function to prepare and collect valid numbers   def   findNDigitNums  (  n  ):   res   =   []   findNDigitNumsUtil  (  0     n     0     0     0     res  )   return   res   # Driver code   if   __name__   ==   '__main__'  :   n   =   2   res   =   findNDigitNums  (  n  )   # Print all collected valid numbers   for   i   in   range  (  len  (  res  )):   print  (  res  [  i  ]   end  =  ' '  )   
C#
   // C# program to print all n-digit numbers such that   // the absolute difference between the sum of digits at   // even and odd positions is 1   using     System  ;   using     System.Collections.Generic  ;   class     GfG     {      // Recursive function to generate numbers      static     void     findNDigitNumsUtil  (  int     pos       int     n       int     num        int     evenSum       int     oddSum        List   <  int  >     res  )     {      // If number is formed      if     (  pos     ==     n  )     {      // Check absolute difference condition      if     (  Math  .  Abs  (  evenSum     -     oddSum  )     ==     1  )     {      res  .  Add  (  num  );      }      return  ;      }      // Digits to consider at current position      for     (  int     d     =     0  ;     d      <=     9  ;     d  ++  )     {      // Skip leading 0      if     (  pos     ==     0     &&     d     ==     0  )     {      continue  ;      }      // If position is even (0-based) add to evenSum      if     (  pos     %     2     ==     0  )     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum     +     d       oddSum       res  );      }      // If position is odd add to oddSum      else     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum       oddSum     +     d       res  );      }      }      }      // Function to prepare and collect valid numbers      static     List   <  int  >     findNDigitNums  (  int     n  )     {      List   <  int  >     res     =     new     List   <  int  >  ();      findNDigitNumsUtil  (  0       n       0       0       0       res  );      return     res  ;      }      // Driver code      public     static     void     Main  (  string  []     args  )     {      int     n     =     2  ;      List   <  int  >     res     =     findNDigitNums  (  n  );      // Print all collected valid numbers      for     (  int     i     =     0  ;     i      <     res  .  Count  ;     i  ++  )     {      Console  .  Write  (  res  [  i  ]     +     ' '  );      }      }   }   
JavaScript
   // JavaScript program to print all n-digit numbers such that   // the absolute difference between the sum of digits at   // even and odd positions is 1   // Recursive function to generate numbers   function     findNDigitNumsUtil  (  pos       n       num       evenSum       oddSum       res  )     {      // If number is formed      if     (  pos     ===     n  )     {      // Check absolute difference condition      if     (  Math  .  abs  (  evenSum     -     oddSum  )     ===     1  )     {      res  .  push  (  num  );      }      return  ;      }      // Digits to consider at current position      for     (  let     d     =     0  ;     d      <=     9  ;     d  ++  )     {      // Skip leading 0      if     (  pos     ===     0     &&     d     ===     0  )     {      continue  ;      }      // If position is even (0-based) add to evenSum      if     (  pos     %     2     ===     0  )     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum     +     d       oddSum       res  );      }      // If position is odd add to oddSum      else     {      findNDigitNumsUtil  (  pos     +     1       n       num     *     10     +     d        evenSum       oddSum     +     d       res  );      }      }   }   // Function to prepare and collect valid numbers   function     findNDigitNums  (  n  )     {      let     res     =     [];      findNDigitNumsUtil  (  0       n       0       0       0       res  );      return     res  ;   }   // Driver code   let     n     =     2  ;   let     res     =     findNDigitNums  (  n  );   // Print all collected valid numbers   for     (  let     i     =     0  ;     i      <     res  .  length  ;     i  ++  )     {      process  .  stdout  .  write  (  res  [  i  ]     +     ' '  );   }   

Sortida
10 12 21 23 32 34 43 45 54 56 65 67 76 78 87 89 98  

Complexitat temporal: O(9 × 10^(n-1)) ja que cada dígit té fins a 10 opcions (excepte la primera que en té 9).
Complexitat espacial: O(n + k) on n és la profunditat de recursivitat i k és el nombre de resultats vàlids.