Najmanjši palindrom po zamenjavi

Podan je niz, ki vsebuje nekaj malih črk in en poseben znak pika (.). Vse pike moramo zamenjati z neko abecedo tako, da nastali niz postane palindrom, v primeru številnih možnih zamenjav moramo izbrati palindromski niz, ki je leksikografsko najmanjši. Če ni mogoče pretvoriti niza v palindrom po vseh možnih zamenjavah, izpiši Ni mogoče. 

Primeri:  

Input : str = ab..e.c.a Output : abcaeacba The smallest palindrome which can be made after replacement is 'abcaeacba' We replaced first dot with 'c' second dot with 'a' third dot with 'a' and fourth dot with 'b' Input : str = ab..e.c.b Output : Not Possible It is not possible to convert above string into palindrome 

To težavo lahko rešimo na naslednji način. Ker mora biti nastali niz palindrom, lahko preverimo par znakov, ki niso pike, pri samem zagonu, če se ne ujemajo, potem neposredna vrnitev ni mogoča, ker lahko nov znak postavimo samo na položaj pik in ne nikamor drugje. 

Nato ponovimo znake niza, če je trenutni znak pika, nato preverimo njegov seznanjeni znak (znak na (n – i -1) mestu), če je tudi ta znak pika, potem lahko oba znaka zamenjamo z 'a', ker je 'a' najmanjša mala črka, kar bo zagotovilo najmanjši leksikografski niz na koncu, zamenjava obeh s katerim koli drugim znakom pa bo povzročila leksikografsko večji palindromski niz. V drugem primeru, če znak v paru ni pika, moramo za izdelavo palindroma niza zamenjati trenutni znak z njegovim znakom v paru. 

So in short If both 'i' and 'n- i- 1' are dot replace them by ‘a’ If one of them is a dot character replace that by other non-dot character 

Z zgornjim postopkom dobimo leksikografsko najmanjši palindromski niz. 

Izvedba:

C++
   // C++ program to get lexicographically smallest   // palindrome string   #include          using     namespace     std  ;   // Utility method to check str is possible palindrome   // after ignoring .   bool     isPossiblePalindrome  (  string     str  )   {      int     n     =     str  .  length  ();      for     (  int     i  =  0  ;     i   <  n  /  2  ;     i  ++  )      {      /* If both left and right character are not    dot and they are not equal also then it    is not possible to make this string a    palindrome */      if     (  str  [  i  ]     !=     '.'     &&      str  [  n  -  i  -1  ]     !=     '.'     &&      str  [  i  ]     !=     str  [  n  -  i  -1  ])      return     false  ;      }      return     true  ;   }   // Returns lexicographically smallest palindrom   // string if possible   string     smallestPalindrome  (  string     str  )   {      if     (  !  isPossiblePalindrome  (  str  ))      return     'Not Possible'  ;      int     n     =     str  .  length  ();      // loop through character of string      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      {      if     (  str  [  i  ]     ==     '.'  )      {      // if one of character is dot replace dot      // with other character      if     (  str  [  n     -     i     -     1  ]     !=     '.'  )      str  [  i  ]     =     str  [  n     -     i     -     1  ];      // if both character are dot then replace      // them with smallest character 'a'      else      str  [  i  ]     =     str  [  n     -     i     -     1  ]     =     'a'  ;      }      }      // return the result      return     str  ;   }   // Driver code to test above methods   int     main  ()   {      string     str     =     'ab..e.c.a'  ;      cout      < <     smallestPalindrome  (  str  )      < <     endl  ;      return     0  ;   }   
Java
   // Java program to get lexicographically    // smallest palindrome string   class   GFG      {   // Utility method to check str is   // possible palindrome after ignoring   static     boolean     isPossiblePalindrome  (  char     str  []  )   {   int     n     =     str  .  length  ;   for     (  int     i     =     0  ;     i      <     n     /     2  ;     i  ++  )   {      /* If both left and right character     are not dot and they are not     equal also then it is not     possible to make this string a    palindrome */      if     (  str  [  i  ]     !=     '.'     &&      str  [  n     -     i     -     1  ]     !=     '.'     &&      str  [  i  ]     !=     str  [  n     -     i     -     1  ]  )      return     false  ;   }   return     true  ;   }   // Returns lexicographically smallest    // palindrome string if possible   static     void     smallestPalindrome  (  char     str  []  )   {   if     (  !  isPossiblePalindrome  (  str  ))      System  .  out  .  println  (  'Not Possible'  );   int     n     =     str  .  length  ;   // loop through character of string   for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )   {      if     (  str  [  i  ]     ==     '.'  )      {      // if one of character is dot       // replace dot with other character      if     (  str  [  n     -     i     -     1  ]     !=     '.'  )      str  [  i  ]     =     str  [  n     -     i     -     1  ]  ;      // if both character are dot       // then replace them with       // smallest character 'a'      else      str  [  i  ]     =     str  [  n     -     i     -     1  ]     =     'a'  ;      }   }   // return the result   for  (  int     i     =     0  ;     i      <     n  ;     i  ++  )      System  .  out  .  print  (  str  [  i  ]     +     ''  );   }   // Driver code   public     static     void     main  (  String  []     args  )   {      String     str     =     'ab..e.c.a'  ;      char  []     s     =     str  .  toCharArray  ();      smallestPalindrome  (  s  );   }   }   // This code is contributed    // by ChitraNayal   
Python 3
   # Python 3 program to get lexicographically    # smallest palindrome string   # Utility method to check str is    # possible palindrome after ignoring    def   isPossiblePalindrome  (  str  ):   n   =   len  (  str  )   for   i   in   range  (  n   //   2  ):   # If both left and right character    # are not dot and they are not    # equal also then it is not possible    # to make this string a palindrome    if   (  str  [  i  ]   !=   '.'   and   str  [  n   -   i   -   1  ]   !=   '.'   and   str  [  i  ]   !=   str  [  n   -   i   -   1  ]):   return   False   return   True   # Returns lexicographically smallest   # palindrome string if possible   def   smallestPalindrome  (  str  ):   if   (  not   isPossiblePalindrome  (  str  )):   return   'Not Possible'   n   =   len  (  str  )   str   =   list  (  str  )   # loop through character of string   for   i   in   range  (  n  ):   if   (  str  [  i  ]   ==   '.'  ):   # if one of character is dot    # replace dot with other character   if   (  str  [  n   -   i   -   1  ]   !=   '.'  ):   str  [  i  ]   =   str  [  n   -   i   -   1  ]   # if both character are dot    # then replace them with    # smallest character 'a'   else  :   str  [  i  ]   =   str  [  n   -   i   -   1  ]   =   'a'   # return the result   return   str   # Driver code   if   __name__   ==   '__main__'  :   str   =   'ab..e.c.a'   print  (  ''  .  join  (  smallestPalindrome  (  str  )))   # This code is contributed by ChitraNayal   
C#
   // C# program to get lexicographically    // smallest palindrome string   using     System  ;   public     class     GFG         {      // Utility method to check str is      // possible palindrome after ignoring      static     bool     isPossiblePalindrome  (  char     []  str  )      {      int     n     =     str  .  Length  ;      for     (  int     i     =     0  ;     i      <     n     /     2  ;     i  ++  )      {      /* If both left and right character     are not dot and they are not     equal also then it is not     possible to make this string a    palindrome */      if     (  str  [  i  ]     !=     '.'     &&      str  [  n     -     i     -     1  ]     !=     '.'     &&      str  [  i  ]     !=     str  [  n     -     i     -     1  ])      return     false  ;      }      return     true  ;      }      // Returns lexicographically smallest       // palindrome string if possible      static     void     smallestPalindrome  (  char     []  str  )      {      if     (  !  isPossiblePalindrome  (  str  ))      Console  .  WriteLine  (  'Not Possible'  );      int     n     =     str  .  Length  ;      // loop through character of string      for     (  int     i     =     0  ;     i      <     n  ;     i  ++  )      {      if     (  str  [  i  ]     ==     '.'  )      {      // if one of character is dot       // replace dot with other character      if     (  str  [  n     -     i     -     1  ]     !=     '.'  )      str  [  i  ]     =     str  [  n     -     i     -     1  ];      // if both character are dot       // then replace them with       // smallest character 'a'      else      str  [  i  ]     =     str  [  n     -     i     -     1  ]     =     'a'  ;      }      }      // return the result      for  (  int     i     =     0  ;     i      <     n  ;     i  ++  )      Console  .  Write  (  str  [  i  ]     +     ''  );      }      // Driver code      public     static     void     Main  ()      {      String     str     =     'ab..e.c.a'  ;      char  []     s     =     str  .  ToCharArray  ();      smallestPalindrome  (  s  );      }   }   // This code is contributed by PrinciRaj1992   
PHP
      // PHP program to get lexicographically   // smallest palindrome string   // Utility method to check str is   // possible palindrome after ignoring   function   isPossiblePalindrome  (  $str  )   {   $n   =   strlen  (  $str  );   for   (  $i   =   0  ;   $i    <   $n   /   2  ;   $i  ++  )   {   /* If both left and right     character are not dot and     they are not equal also then     it is not possible to make this     string a palindrome */   if   (  $str  [  $i  ]   !=   '.'   &&   $str  [  $n   -   $i   -   1  ]   !=   '.'   &&   $str  [  $i  ]   !=   $str  [  $n   -   $i   -   1  ])   return   false  ;   }   return   true  ;   }   // Returns lexicographically smallest    // palindrome string if possible   function   smallestPalindrome  (  $str  )   {   if   (  !  isPossiblePalindrome  (  $str  ))   return   'Not Possible'  ;   $n   =   strlen  (  $str  );   // loop through character of string   for   (  $i  =   0  ;   $i    <   $n  ;   $i  ++  )   {   if   (  $str  [  $i  ]   ==   '.'  )   {   // if one of character is dot    // replace dot with other character   if   (  $str  [  $n   -   $i   -   1  ]   !=   '.'  )   $str  [  $i  ]   =   $str  [  $n   -   $i   -   1  ];   // if both character are dot    // then replace them with    // smallest character 'a'   else   $str  [  $i  ]   =   $str  [  $n   -   $i   -   1  ]   =   'a'  ;   }   }   // return the result   return   $str  ;   }   // Driver code   $str   =   'ab..e.c.a'  ;   echo   smallestPalindrome  (  $str  );   // This code is contributed    // by ChitraNayal   ?>   
JavaScript
    <  script  >   // Javascript program to get lexicographically    // smallest palindrome string          // Utility method to check str is      // possible palindrome after ignoring      function     isPossiblePalindrome  (  str  )      {      let     n     =     str  .  length  ;      for     (  let     i     =     0  ;     i      <     Math  .  floor  (  n     /     2  );     i  ++  )      {      /* If both left and right character     are not dot and they are not     equal also then it is not     possible to make this string a    palindrome */      if     (  str  [  i  ]     !=     '.'     &&      str  [  n     -     i     -     1  ]     !=     '.'     &&      str  [  i  ]     !=     str  [  n     -     i     -     1  ])      return     false  ;      }          return     true  ;      }          // Returns lexicographically smallest       // palindrome string if possible      function     smallestPalindrome  (  str  )      {      if     (  !  isPossiblePalindrome  (  str  ))      document  .  write  (  'Not Possible'  );          let     n     =     str  .  length  ;          // loop through character of string      for     (  let     i     =     0  ;     i      <     n  ;     i  ++  )      {      if     (  str  [  i  ]     ==     '.'  )      {      // if one of character is dot       // replace dot with other character      if     (  str  [  n     -     i     -     1  ]     !=     '.'  )      str  [  i  ]     =     str  [  n     -     i     -     1  ];          // if both character are dot       // then replace them with       // smallest character 'a'      else      str  [  i  ]     =     str  [  n     -     i     -     1  ]     =     'a'  ;      }      }          // return the result      for  (  let     i     =     0  ;     i      <     n  ;     i  ++  )      document  .  write  (  str  [  i  ]     +     ''  );          }          // Driver code      let     str  =  'ab..e.c.a'  ;      let     s     =     str  .  split  (  ''  );      smallestPalindrome  (  s  );          // This code is contributed by rag2127        <  /script>   

Izhod
abcaeacba 

Časovna zahtevnost: O(n) kjer je n dolžina niza.
Kompleksnost pomožnega prostora: O(1)