Odstraňte znak z řetězce, abyste z něj udělali palindrom

Daný řetězec potřebujeme zkontrolovat, zda je možné z tohoto řetězce udělat palindrom poté, co z něj odstraníme právě jeden znak. 

Příklady:   

Input : str = abcba Output : Yes we can remove character ‘c’ to make string palindrome Input : str = abcbea Output : Yes we can remove character ‘e’ to make string palindrome Input : str = abecbea It is not possible to make this string palindrome just by removing one character  

Tento problém můžeme vyřešit nalezením pozice nesouladu. Začneme opakovat řetězec tak, že na obou koncích ponecháme dva ukazatele, které se po každé iteraci pohybují směrem ke střední poloze, tato iterace se zastaví, když najdeme nesoulad, protože je povoleno odstranit pouze jeden znak, zde máme dvě možnosti

Při neshodě buď odeberte znak, na který ukazuje levý ukazatel, nebo odeberte znak, na který ukazuje pravý ukazatel.

Zkontrolujeme oba případy, které si pamatujeme, protože jsme prošli stejný počet kroků z obou stran, tento střední řetězec by měl být také palindrom po odstranění jednoho znaku, takže zkontrolujeme dva podřetězce jeden odstraněním levého znaku a jeden odstraněním pravého znaku a pokud jeden z nich je palindrom, můžeme vytvořit úplný palindrom řetězce odstraněním odpovídajícího znaku a pokud oba podřetězce nejsou pod daným řetězcem palindrom, není možné provést palindro pod daným řetězcem. 

Implementace:

C++
   // C/C++ program to check whether it is possible to make   // string palindrome by removing one character   #include          using     namespace     std  ;   // Utility method to check if substring from low to high is   // palindrome or not.   bool     isPalindrome  (  string  ::  iterator     low       string  ::  iterator     high  )   {      while     (  low      <     high  )      {      if     (  *  low     !=     *  high  )      return     false  ;      low  ++  ;      high  --  ;      }      return     true  ;   }   // This method returns -1 if it is not possible to make string   // a palindrome. It returns -2 if string is already a palindrome.   // Otherwise it returns index of character whose removal can   // make the whole string palindrome.   int     possiblePalinByRemovingOneChar  (  string     str  )   {      // Initialize low and high by both the ends of the string      int     low     =     0       high     =     str  .  length  ()     -     1  ;      // loop until low and high cross each other      while     (  low      <     high  )      {      // If both characters are equal then move both pointer      // towards end      if     (  str  [  low  ]     ==     str  [  high  ])      {      low  ++  ;      high  --  ;      }      else      {      /* If removing str[low] makes the whole string palindrome.    We basically check if substring str[low+1..high] is    palindrome or not. */      if     (  isPalindrome  (  str  .  begin  ()     +     low     +     1       str  .  begin  ()     +     high  ))      return     low  ;      /* If removing str[high] makes the whole string palindrome    We basically check if substring str[low+1..high] is    palindrome or not. */      if     (  isPalindrome  (  str  .  begin  ()     +     low       str  .  begin  ()     +     high     -     1  ))      return     high  ;      return     -1  ;      }      }      // We reach here when complete string will be palindrome      // if complete string is palindrome then return mid character      return     -2  ;   }   // Driver code to test above methods   int     main  ()   {      string     str     =     'abecbea'  ;      int     idx     =     possiblePalinByRemovingOneChar  (  str  );      if     (  idx     ==     -1  )      cout      < <     'Not Possible   n  '  ;      else     if     (  idx     ==     -2  )      cout      < <     'Possible without removing any character'  ;      else      cout      < <     'Possible by removing character'       < <     ' at index '      < <     idx      < <     '  n  '  ;      return     0  ;   }   
Java
   // Java program to check whether    // it is possible to make string    // palindrome by removing one character   import     java.util.*  ;   class   GFG      {      // Utility method to check if       // substring from low to high is      // palindrome or not.      static     boolean     isPalindrome  (  String     str           int     low       int     high  )      {      while     (  low      <     high  )         {      if     (  str  .  charAt  (  low  )     !=     str  .  charAt  (  high  ))      return     false  ;      low  ++  ;      high  --  ;      }      return     true  ;      }      // This method returns -1 if it is       // not possible to make string a palindrome.       // It returns -2 if string is already       // a palindrome. Otherwise it returns       // index of character whose removal can      // make the whole string palindrome.      static     int     possiblePalinByRemovingOneChar  (  String     str  )      {      // Initialize low and right       // by both the ends of the string      int     low     =     0       high     =     str  .  length  ()     -     1  ;      // loop until low and      // high cross each other      while     (  low      <     high  )         {      // If both characters are equal then       // move both pointer towards end      if     (  str  .  charAt  (  low  )     ==     str  .  charAt  (  high  ))         {      low  ++  ;      high  --  ;      }         else      {      /*    * If removing str[low] makes the     * whole string palindrome. We basically     * check if substring str[low+1..high]    * is palindrome or not.    */      if     (  isPalindrome  (  str       low     +     1       high  ))      return     low  ;      /*    * If removing str[high] makes the whole string     * palindrome. We basically check if substring     * str[low+1..high] is palindrome or not.    */      if     (  isPalindrome  (  str       low       high     -     1  ))      return     high  ;      return     -  1  ;      }      }      // We reach here when complete string       // will be palindrome if complete string       // is palindrome then return mid character      return     -  2  ;      }      // Driver Code      public     static     void     main  (  String  []     args  )      {      String     str     =     'abecbea'  ;      int     idx     =     possiblePalinByRemovingOneChar  (  str  );      if     (  idx     ==     -  1  )      System  .  out  .  println  (  'Not Possible'  );      else     if     (  idx     ==     -  2  )      System  .  out  .  println  (  'Possible without '     +         'removing any character'  );      else      System  .  out  .  println  (  'Possible by removing'     +         ' character at index '     +     idx  );      }   }   // This code is contributed by   // sanjeev2552   
Python3
   # Python program to check whether it is possible to make   # string palindrome by removing one character   # Utility method to check if substring from    # low to high is palindrome or not.   def   isPalindrome  (  string  :   str     low  :   int     high  :   int  )   ->   bool  :   while   low    <   high  :   if   string  [  low  ]   !=   string  [  high  ]:   return   False   low   +=   1   high   -=   1   return   True   # This method returns -1 if it    # is not possible to make string   # a palindrome. It returns -2 if    # string is already a palindrome.   # Otherwise it returns index of   # character whose removal can   # make the whole string palindrome.   def   possiblepalinByRemovingOneChar  (  string  :   str  )   ->   int  :   # Initialize low and right by   # both the ends of the string   low   =   0   high   =   len  (  string  )   -   1   # loop until low and high cross each other   while   low    <   high  :   # If both characters are equal then   # move both pointer towards end   if   string  [  low  ]   ==   string  [  high  ]:   low   +=   1   high   -=   1   else  :   # If removing str[low] makes the whole string palindrome.   # We basically check if substring str[low+1..high] is   # palindrome or not.   if   isPalindrome  (  string     low   +   1     high  ):   return   low   # If removing str[high] makes the whole string palindrome   # We basically check if substring str[low+1..high] is   # palindrome or not   if   isPalindrome  (  string     low     high   -   1  ):   return   high   return   -  1   # We reach here when complete string will be palindrome   # if complete string is palindrome then return mid character   return   -  2   # Driver Code   if   __name__   ==   '__main__'  :   string   =   'abecbea'   idx   =   possiblepalinByRemovingOneChar  (  string  )   if   idx   ==   -  1  :   print  (  'Not possible'  )   else   if   idx   ==   -  2  :   print  (  'Possible without removing any character'  )   else  :   print  (  'Possible by removing character at index'     idx  )   # This code is contributed by   # sanjeev2552   
C#
   // C# program to check whether    // it is possible to make string    // palindrome by removing one character   using     System  ;   class     GFG      {      // Utility method to check if       // substring from low to high is      // palindrome or not.      static     bool     isPalindrome  (  string     str       int     low       int     high  )      {      while     (  low      <     high  )         {      if     (  str  [  low  ]     !=     str  [  high  ])      return     false  ;      low  ++  ;      high  --  ;      }      return     true  ;      }      // This method returns -1 if it is       // not possible to make string a palindrome.       // It returns -2 if string is already       // a palindrome. Otherwise it returns       // index of character whose removal can      // make the whole string palindrome.      static     int     possiblePalinByRemovingOneChar  (  string     str  )      {      // Initialize low and right       // by both the ends of the string      int     low     =     0       high     =     str  .  Length     -     1  ;      // loop until low and      // high cross each other      while     (  low      <     high  )         {      // If both characters are equal then       // move both pointer towards end      if     (  str  [  low  ]     ==     str  [  high  ])         {      low  ++  ;      high  --  ;      }         else      {      /*    * If removing str[low] makes the     * whole string palindrome. We basically     * check if substring str[low+1..high]    * is palindrome or not.    */      if     (  isPalindrome  (  str       low     +     1       high  ))      return     low  ;      /*    * If removing str[high] makes the whole string     * palindrome. We basically check if substring     * str[low+1..high] is palindrome or not.    */      if     (  isPalindrome  (  str       low       high     -     1  ))      return     high  ;      return     -  1  ;      }      }      // We reach here when complete string       // will be palindrome if complete string       // is palindrome then return mid character      return     -  2  ;      }      // Driver Code      public     static     void     Main  (  String  []     args  )      {      string     str     =     'abecbea'  ;      int     idx     =     possiblePalinByRemovingOneChar  (  str  );      if     (  idx     ==     -  1  )      Console  .  Write  (  'Not Possible'  );      else     if     (  idx     ==     -  2  )      Console  .  Write  (  'Possible without '     +         'removing any character'  );      else      Console  .  Write  (  'Possible by removing'     +         ' character at index '     +     idx  );      }   }   // This code is contributed by shivanisinghss2110   
JavaScript
    <  script  >   // JavaScript program to check whether    // it is possible to make string    // palindrome by removing one character   // Utility method to check if    // substring from low to high is   // palindrome or not.   function     isPalindrome  (  str       low       high  )      {      while     (  low      <     high  )         {      if     (  str  .  charAt  (  low  )     !=     str  .  charAt  (  high  ))      return     false  ;      low  ++  ;      high  --  ;      }      return     true  ;      }      // This method returns -1 if it is       // not possible to make string a palindrome.       // It returns -2 if string is already       // a palindrome. Otherwise it returns       // index of character whose removal can      // make the whole string palindrome.      function     possiblePalinByRemovingOneChar  (  str  )      {      // Initialize low and right       // by both the ends of the string      var     low     =     0       high     =     str  .  length     -     1  ;      // loop until low and      // high cross each other      while     (  low      <     high  )         {      // If both characters are equal then       // move both pointer towards end      if     (  str  .  charAt  (  low  )     ==     str  .  charAt  (  high  ))         {      low  ++  ;      high  --  ;      }         else      {      /*    * If removing str[low] makes the     * whole string palindrome. We basically     * check if substring str[low+1..high]    * is palindrome or not.    */      if     (  isPalindrome  (  str       low     +     1       high  ))      return     low  ;      /*    * If removing str[high] makes the whole string     * palindrome. We basically check if substring     * str[low+1..high] is palindrome or not.    */      if     (  isPalindrome  (  str       low       high     -     1  ))      return     high  ;      return     -  1  ;      }      }      // We reach here when complete string       // will be palindrome if complete string       // is palindrome then return mid character      return     -  2  ;      }      // Driver Code      var     str     =     'abecbea'  ;      var     idx     =     possiblePalinByRemovingOneChar  (  str  );      if     (  idx     ==     -  1  )      document  .  write  (  'Not Possible'  );      else     if     (  idx     ==     -  2  )      document  .  write  (  'Possible without '     +         'removing any character'  );      else      document  .  write  (  'Possible by removing'     +         ' character at index '     +     idx  );   // this code is contributed by shivanisinghss2110    <  /script>   

Výstup
Not Possible  

Časová složitost: O(N)
Vesmírná složitost: O(1)

 

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