Palindrome عن طريق الإدراج الأمامي

Palindrome عن طريق الإدراج الأمامي
جربه على ممارسة GfG الأحرف المراد إضافتها في المقدمة لـ Palindrome

بالنظر إلى سلسلة تتكون من أحرف إنجليزية صغيرة فقط، ابحث عن الحد الأدنى عدد الحروف التي يجب أن تكون وأضاف إلى أمام من الصورة لجعلها متناظرة.
ملحوظة: المتناظر هو سلسلة تُقرأ بنفس الطريقة للأمام والخلف.

أمثلة:  

مدخل : ق = 'اي بي سي'
الإخراج : 2
توضيح : يمكننا إنشاء متناظرة السلسلة أعلاه كـ "cbabc" عن طريق إضافة "b" و "c" في المقدمة.

مدخل : ق = 'aacecaaaa'
الإخراج : 2
توضيح : يمكننا أن نجعل متناظرة السلسلة أعلاه كـ "aaaacecaaaa" عن طريق إضافة اثنين من الحروف في مقدمة السلسلة.

جدول المحتويات

[نهج ساذج] التحقق من جميع البادئات - O(n^2) Time وO(1) Space

تعتمد الفكرة على الملاحظة التي مفادها أننا نحتاج إلى العثور على أطول بادئة من سلسلة معينة والتي تعد أيضًا متناظرة. ثم الحد الأدنى من الأحرف الأمامية المراد إضافتها لإنشاء سلسلة متناظرة معينة سيكون هو الأحرف المتبقية.

C++
   #include          using     namespace     std  ;   // function to check if the substring s[i...j] is a palindrome   bool     isPalindrome  (  string     &  s       int     i       int     j  )     {      while     (  i      <     j  )     {          // if characters at the ends are not equal       // it's not a palindrome      if     (  s  [  i  ]     !=     s  [  j  ])     {      return     false  ;      }      i  ++  ;      j  --  ;      }      return     true  ;   }   int     minChar  (  string     &  s  )     {      int     cnt     =     0  ;      int     i     =     s  .  size  ()     -     1  ;          // iterate from the end of the string checking for the       // longestpalindrome starting from the beginning      while     (  i     >=     0     &&     !  isPalindrome  (  s       0       i  ))     {          i  --  ;      cnt  ++  ;      }          return     cnt  ;   }   int     main  ()     {      string     s     =     'aacecaaaa'  ;      cout      < <     minChar  (  s  );      return     0  ;   }   
C
   #include         #include         #include         // function to check if the substring s[i...j] is a palindrome   bool     isPalindrome  (  char     s  []     int     i       int     j  )     {      while     (  i      <     j  )     {          // if characters at the ends are not the same       // it's not a palindrome      if     (  s  [  i  ]     !=     s  [  j  ])     {      return     false  ;      }      i  ++  ;      j  --  ;      }      return     true  ;   }   int     minChar  (  char     s  [])     {      int     cnt     =     0  ;      int     i     =     strlen  (  s  )     -     1  ;          // iterate from the end of the string checking for the       // longest palindrome starting from the beginning      while     (  i     >=     0     &&     !  isPalindrome  (  s       0       i  ))     {          i  --  ;      cnt  ++  ;      }          return     cnt  ;   }   int     main  ()     {          char     s  []     =     'aacecaaaa'  ;      printf  (  '%d'       minChar  (  s  ));      return     0  ;   }   
Java
   class   GfG     {      // function to check if the substring       // s[i...j] is a palindrome      static     boolean     isPalindrome  (  String     s       int     i       int     j  )     {      while     (  i      <     j  )     {          // if characters at the ends are not the same       // it's not a palindrome      if     (  s  .  charAt  (  i  )     !=     s  .  charAt  (  j  ))     {      return     false  ;      }      i  ++  ;      j  --  ;      }      return     true  ;      }      static     int     minChar  (  String     s  )     {      int     cnt     =     0  ;      int     i     =     s  .  length  ()     -     1  ;          // iterate from the end of the string checking for the       // longest palindrome starting from the beginning      while     (  i     >=     0     &&     !  isPalindrome  (  s       0       i  ))     {      i  --  ;      cnt  ++  ;      }          return     cnt  ;      }      public     static     void     main  (  String  []     args  )     {      String     s     =     'aacecaaaa'  ;      System  .  out  .  println  (  minChar  (  s  ));      }   }   
Python
   # function to check if the substring s[i...j] is a palindrome   def   isPalindrome  (  s     i     j  ):   while   i    <   j  :   # if characters at the ends are not the same    # it's not a palindrome   if   s  [  i  ]   !=   s  [  j  ]:   return   False   i   +=   1   j   -=   1   return   True   def   minChar  (  s  ):   cnt   =   0   i   =   len  (  s  )   -   1   # iterate from the end of the string checking for the    # longest palindrome starting from the beginning   while   i   >=   0   and   not   isPalindrome  (  s     0     i  ):   i   -=   1   cnt   +=   1   return   cnt   if   __name__   ==   '__main__'  :   s   =   'aacecaaaa'   print  (  minChar  (  s  ))   
C#
   using     System  ;   class     GfG     {      // function to check if the substring s[i...j] is a palindrome      static     bool     isPalindrome  (  string     s       int     i       int     j  )     {      while     (  i      <     j  )     {          // if characters at the ends are not the same       // it's not a palindrome      if     (  s  [  i  ]     !=     s  [  j  ])     {      return     false  ;      }      i  ++  ;      j  --  ;      }      return     true  ;      }      static     int     minChar  (  string     s  )     {      int     cnt     =     0  ;      int     i     =     s  .  Length     -     1  ;          // iterate from the end of the string checking for the longest       // palindrome starting from the beginning      while     (  i     >=     0     &&     !  isPalindrome  (  s       0       i  ))     {      i  --  ;      cnt  ++  ;      }          return     cnt  ;      }      static     void     Main  ()     {          string     s     =     'aacecaaaa'  ;      Console  .  WriteLine  (  minChar  (  s  ));      }   }   
JavaScript
   // function to check if the substring s[i...j] is a palindrome   function     isPalindrome  (  s       i       j  )     {      while     (  i      <     j  )     {          // if characters at the ends are not the same       // it's not a palindrome      if     (  s  [  i  ]     !==     s  [  j  ])     {      return     false  ;      }      i  ++  ;      j  --  ;      }      return     true  ;   }   function     minChar  (  s  )     {      let     cnt     =     0  ;      let     i     =     s  .  length     -     1  ;          // iterate from the end of the string checking for the      // longest palindrome starting from the beginning      while     (  i     >=     0     &&     !  isPalindrome  (  s       0       i  ))     {          i  --  ;      cnt  ++  ;      }          return     cnt  ;   }   // Driver code   let     s     =     'aacecaaaa'  ;   console  .  log  (  minChar  (  s  ));   

الإخراج
2 

[النهج المتوقع 1] استخدام مصفوفة lps لخوارزمية KMP - وقت O(n) ومسافة O(n)

الملاحظة الأساسية هي أن أطول بادئة متناوبة لسلسلة ما تصبح أطول لاحقة متناوبة على عكسها.
بالنظر إلى سلسلة s = 'aacecaaaa'، فإن revS العكسي = 'aaaacecaa'. أطول بادئة متناوبة لـ s هي "aacecaa".

للعثور على هذا بكفاءة نستخدم مجموعة LPS من خوارزمية KMP . نقوم بربط السلسلة الأصلية بحرف خاص وعكسه: s + '$' + revS.
تساعد مصفوفة LPS لهذه السلسلة المدمجة في تحديد البادئة الأطول لـ s التي تتطابق مع لاحقة revS والتي تمثل أيضًا البادئة المتناوبة لـ s.

تخبرنا القيمة الأخيرة لمصفوفة LPS بعدد الأحرف التي تشكل بالفعل متناظرًا في البداية. وبالتالي فإن الحد الأدنى لعدد الأحرف التي يجب إضافتها لجعل s متناظرًا هو s.length() - lps.back().

C++
   #include          #include          #include         using     namespace     std  ;   vector   <  int  >     computeLPSArray  (  string     &  pat  )     {      int     n     =     pat  .  length  ();      vector   <  int  >     lps  (  n  );      // lps[0] is always 0      lps  [  0  ]     =     0  ;      int     len     =     0  ;      // loop calculates lps[i] for i = 1 to M-1      int     i     =     1  ;      while     (  i      <     n  )     {      // if the characters match increment len      // and set lps[i]      if     (  pat  [  i  ]     ==     pat  [  len  ])     {      len  ++  ;      lps  [  i  ]     =     len  ;      i  ++  ;      }      // if there is a mismatch      else     {      // if len is not zero update len to      // the last known prefix length      if     (  len     !=     0  )     {      len     =     lps  [  len     -     1  ];      }      // no prefix matches set lps[i] to 0      else     {      lps  [  i  ]     =     0  ;      i  ++  ;      }      }      }      return     lps  ;   }   // returns minimum character to be added at   // front to make string palindrome   int     minChar  (  string     &  s  )     {      int     n     =     s  .  length  ();      string     rev     =     s  ;      reverse  (  rev  .  begin  ()     rev  .  end  ());      // get concatenation of string special character      // and reverse string      s     =     s     +     '$'     +     rev  ;      // get LPS array of this concatenated string      vector   <  int  >     lps     =     computeLPSArray  (  s  );      // by subtracting last entry of lps vector from      // string length we will get our result      return     (  n     -     lps  .  back  ());   }   int     main  ()     {      string     s     =     'aacecaaaa'  ;      cout      < <     minChar  (  s  );      return     0  ;   }   
Java
   import     java.util.ArrayList  ;   class   GfG     {      static     int  []     computeLPSArray  (  String     pat  )     {      int     n     =     pat  .  length  ();      int  []     lps     =     new     int  [  n  ]  ;      // lps[0] is always 0      lps  [  0  ]     =     0  ;      int     len     =     0  ;      // loop calculates lps[i] for i = 1 to n-1      int     i     =     1  ;      while     (  i      <     n  )     {      // if the characters match increment len      // and set lps[i]      if     (  pat  .  charAt  (  i  )     ==     pat  .  charAt  (  len  ))     {      len  ++  ;      lps  [  i  ]     =     len  ;      i  ++  ;      }      // if there is a mismatch      else     {      // if len is not zero update len to      // the last known prefix length      if     (  len     !=     0  )     {      len     =     lps  [  len     -     1  ]  ;      }      // no prefix matches set lps[i] to 0      else     {      lps  [  i  ]     =     0  ;      i  ++  ;      }      }      }      return     lps  ;      }      // returns minimum character to be added at      // front to make string palindrome      static     int     minChar  (  String     s  )     {      int     n     =     s  .  length  ();      String     rev      =     new     StringBuilder  (  s  ).  reverse  ().  toString  ();      // get concatenation of string special character      // and reverse string      s     =     s     +     '$'     +     rev  ;      // get LPS array of this concatenated string      int  []     lps     =     computeLPSArray  (  s  );      // by subtracting last entry of lps array from      // string length we will get our result      return     (  n     -     lps  [  lps  .  length     -     1  ]  );      }      public     static     void     main  (  String  []     args  )     {      String     s     =     'aacecaaaa'  ;      System  .  out  .  println  (  minChar  (  s  ));      }   }   
Python
   def   computeLPSArray  (  pat  ):   n   =   len  (  pat  )   lps   =   [  0  ]   *   n   # lps[0] is always 0   len_lps   =   0   # loop calculates lps[i] for i = 1 to n-1   i   =   1   while   i    <   n  :   # if the characters match increment len   # and set lps[i]   if   pat  [  i  ]   ==   pat  [  len_lps  ]:   len_lps   +=   1   lps  [  i  ]   =   len_lps   i   +=   1   # if there is a mismatch   else  :   # if len is not zero update len to    # the last known prefix length   if   len_lps   !=   0  :   len_lps   =   lps  [  len_lps   -   1  ]   # no prefix matches set lps[i] to 0   else  :   lps  [  i  ]   =   0   i   +=   1   return   lps   # returns minimum character to be added at   # front to make string palindrome   def   minChar  (  s  ):   n   =   len  (  s  )   rev   =   s  [::  -  1  ]   # get concatenation of string special character   # and reverse string   s   =   s   +   '$'   +   rev   # get LPS array of this concatenated string   lps   =   computeLPSArray  (  s  )   # by subtracting last entry of lps array from   # string length we will get our result   return   n   -   lps  [  -  1  ]   if   __name__   ==   '__main__'  :   s   =   'aacecaaaa'   print  (  minChar  (  s  ))   
C#
   using     System  ;   class     GfG     {      static     int  []     computeLPSArray  (  string     pat  )     {      int     n     =     pat  .  Length  ;      int  []     lps     =     new     int  [  n  ];      // lps[0] is always 0      lps  [  0  ]     =     0  ;      int     len     =     0  ;      // loop calculates lps[i] for i = 1 to n-1      int     i     =     1  ;      while     (  i      <     n  )     {      // if the characters match increment len      // and set lps[i]      if     (  pat  [  i  ]     ==     pat  [  len  ])     {      len  ++  ;      lps  [  i  ]     =     len  ;      i  ++  ;      }      // if there is a mismatch      else     {      // if len is not zero update len to      // the last known prefix length      if     (  len     !=     0  )     {      len     =     lps  [  len     -     1  ];      }      // no prefix matches set lps[i] to 0      else     {      lps  [  i  ]     =     0  ;      i  ++  ;      }      }      }      return     lps  ;      }      // minimum character to be added at      // front to make string palindrome      static     int     minChar  (  string     s  )     {      int     n     =     s  .  Length  ;      char  []     charArray     =     s  .  ToCharArray  ();      Array  .  Reverse  (  charArray  );      string     rev     =     new     string  (  charArray  );      // get concatenation of string special character      // and reverse string      s     =     s     +     '$'     +     rev  ;      // get LPS array of this concatenated string      int  []     lps     =     computeLPSArray  (  s  );      // by subtracting last entry of lps array from      // string length we will get our result      return     n     -     lps  [  lps  .  Length     -     1  ];      }      static     void     Main  ()     {      string     s     =     'aacecaaaa'  ;      Console  .  WriteLine  (  minChar  (  s  ));      }   }   
JavaScript
   function     computeLPSArray  (  pat  )     {      let     n     =     pat  .  length  ;      let     lps     =     new     Array  (  n  ).  fill  (  0  );      // lps[0] is always 0      let     len     =     0  ;      // loop calculates lps[i] for i = 1 to n-1      let     i     =     1  ;      while     (  i      <     n  )     {      // if the characters match increment len      // and set lps[i]      if     (  pat  [  i  ]     ===     pat  [  len  ])     {      len  ++  ;      lps  [  i  ]     =     len  ;      i  ++  ;      }      // if there is a mismatch      else     {      // if len is not zero update len to      // the last known prefix length      if     (  len     !==     0  )     {      len     =     lps  [  len     -     1  ];      }      // no prefix matches set lps[i] to 0      else     {      lps  [  i  ]     =     0  ;      i  ++  ;      }      }      }      return     lps  ;   }   // returns minimum character to be added at   // front to make string palindrome   function     minChar  (  s  )     {      let     n     =     s  .  length  ;      let     rev     =     s  .  split  (  ''  ).  reverse  ().  join  (  ''  );      // get concatenation of string special character      // and reverse string      s     =     s     +     '$'     +     rev  ;      // get LPS array of this concatenated string      let     lps     =     computeLPSArray  (  s  );      // by subtracting last entry of lps array from      // string length we will get our result      return     n     -     lps  [  lps  .  length     -     1  ];   }   // Driver Code   let     s     =     'aacecaaaa'  ;   console  .  log  (  minChar  (  s  ));   

الإخراج
2 

[النهج المتوقع 2] استخدام خوارزمية ماناشر

الفكرة هي الاستخدام خوارزمية ماناشر للعثور بكفاءة على جميع السلاسل الفرعية المتناوبة في الوقت الخطي.
نقوم بتحويل السلسلة عن طريق إدخال أحرف خاصة (#) للتعامل مع متناظرات الطول الزوجية والفردية بشكل موحد.
بعد المعالجة المسبقة، نقوم بالمسح الضوئي من نهاية السلسلة الأصلية ونستخدم مصفوفة نصف قطر المتناظر للتحقق مما إذا كانت البادئة s[0...i] متناظرة. أول فهرس من هذا القبيل i يعطينا أطول بادئة متناوبة ونعيد n - (i + 1) كحد أدنى من الأحرف المراد إضافتها.

C++
   #include          #include         #include         using     namespace     std  ;   // manacher's algorithm for finding longest    // palindromic substrings   class     manacher     {   public  :      // array to store palindrome lengths centered       // at each position      vector   <  int  >     p  ;      // modified string with separators and sentinels      string     ms  ;         manacher  (  string     &  s  )     {      ms     =     '@'  ;      for     (  char     c     :     s  )     {      ms     +=     '#'     +     string  (  1       c  );      }      ms     +=     '#$'  ;      runManacher  ();      }      // core Manacher's algorithm      void     runManacher  ()     {      int     n     =     ms  .  size  ();      p  .  assign  (  n       0  );      int     l     =     0       r     =     0  ;      for     (  int     i     =     1  ;     i      <     n     -     1  ;     ++  i  )     {      if     (  i      <     r  )      p  [  i  ]     =     min  (  r     -     i       p  [  r     +     l     -     i  ]);      // expand around the current center      while     (  ms  [  i     +     1     +     p  [  i  ]]     ==     ms  [  i     -     1     -     p  [  i  ]])      ++  p  [  i  ];      // update center if palindrome goes beyond      // current right boundary      if     (  i     +     p  [  i  ]     >     r  )     {      l     =     i     -     p  [  i  ];      r     =     i     +     p  [  i  ];      }      }      }      // returns the length of the longest palindrome      // centered at given position      int     getLongest  (  int     cen       int     odd  )     {      int     pos     =     2     *     cen     +     2     +     !  odd  ;      return     p  [  pos  ];      }      // checks whether substring s[l...r] is a palindrome      bool     check  (  int     l       int     r  )     {      int     len     =     r     -     l     +     1  ;      int     longest     =     getLongest  ((  l     +     r  )     /     2       len     %     2  );      return     len      <=     longest  ;      }   };   // returns the minimum number of characters to add at the    // front to make the given string a palindrome   int     minChar  (  string     &  s  )     {      int     n     =     s  .  size  ();      manacher     m  (  s  );      // scan from the end to find the longest       // palindromic prefix      for     (  int     i     =     n     -     1  ;     i     >=     0  ;     --  i  )     {      if     (  m  .  check  (  0       i  ))      return     n     -     (  i     +     1  );      }      return     n     -     1  ;   }   int     main  ()     {      string     s     =     'aacecaaaa'  ;      cout      < <     minChar  (  s  )      < <     endl  ;      return     0  ;   }   
Java
   class   GfG     {          // manacher's algorithm for finding longest       // palindromic substrings      static     class   manacher     {      // array to store palindrome lengths centered       // at each position      int  []     p  ;      // modified string with separators and sentinels      String     ms  ;      manacher  (  String     s  )     {      StringBuilder     sb     =     new     StringBuilder  (  '@'  );      for     (  char     c     :     s  .  toCharArray  ())     {      sb  .  append  (  '#'  ).  append  (  c  );      }      sb  .  append  (  '#$'  );      ms     =     sb  .  toString  ();      runManacher  ();      }      // core Manacher's algorithm      void     runManacher  ()     {      int     n     =     ms  .  length  ();      p     =     new     int  [  n  ]  ;      int     l     =     0       r     =     0  ;      for     (  int     i     =     1  ;     i      <     n     -     1  ;     ++  i  )     {      if     (  i      <     r  )      p  [  i  ]     =     Math  .  min  (  r     -     i       p  [  r     +     l     -     i  ]  );      // expand around the current center      while     (  ms  .  charAt  (  i     +     1     +     p  [  i  ]  )     ==     ms  .  charAt  (  i     -     1     -     p  [  i  ]  ))      p  [  i  ]++  ;      // update center if palindrome goes beyond       // current right boundary      if     (  i     +     p  [  i  ]     >     r  )     {      l     =     i     -     p  [  i  ]  ;      r     =     i     +     p  [  i  ]  ;      }      }      }      // returns the length of the longest palindrome       // centered at given position      int     getLongest  (  int     cen       int     odd  )     {      int     pos     =     2     *     cen     +     2     +     (  odd     ==     0     ?     1     :     0  );      return     p  [  pos  ]  ;      }      // checks whether substring s[l...r] is a palindrome      boolean     check  (  int     l       int     r  )     {      int     len     =     r     -     l     +     1  ;      int     longest     =     getLongest  ((  l     +     r  )     /     2       len     %     2  );      return     len      <=     longest  ;      }      }      // returns the minimum number of characters to add at the       // front to make the given string a palindrome      static     int     minChar  (  String     s  )     {      int     n     =     s  .  length  ();      manacher     m     =     new     manacher  (  s  );      // scan from the end to find the longest       // palindromic prefix      for     (  int     i     =     n     -     1  ;     i     >=     0  ;     --  i  )     {      if     (  m  .  check  (  0       i  ))      return     n     -     (  i     +     1  );      }      return     n     -     1  ;      }      public     static     void     main  (  String  []     args  )     {      String     s     =     'aacecaaaa'  ;      System  .  out  .  println  (  minChar  (  s  ));      }   }   
Python
   # manacher's algorithm for finding longest    # palindromic substrings   class   manacher  :   # array to store palindrome lengths centered    # at each position   def   __init__  (  self     s  ):   # modified string with separators and sentinels   self  .  ms   =   '@'   for   c   in   s  :   self  .  ms   +=   '#'   +   c   self  .  ms   +=   '#$'   self  .  p   =   []   self  .  runManacher  ()   # core Manacher's algorithm   def   runManacher  (  self  ):   n   =   len  (  self  .  ms  )   self  .  p   =   [  0  ]   *   n   l   =   r   =   0   for   i   in   range  (  1     n   -   1  ):   if   i    <   r  :   self  .  p  [  i  ]   =   min  (  r   -   i     self  .  p  [  r   +   l   -   i  ])   # expand around the current center   while   self  .  ms  [  i   +   1   +   self  .  p  [  i  ]]   ==   self  .  ms  [  i   -   1   -   self  .  p  [  i  ]]:   self  .  p  [  i  ]   +=   1   # update center if palindrome goes beyond    # current right boundary   if   i   +   self  .  p  [  i  ]   >   r  :   l   =   i   -   self  .  p  [  i  ]   r   =   i   +   self  .  p  [  i  ]   # returns the length of the longest palindrome    # centered at given position   def   getLongest  (  self     cen     odd  ):   pos   =   2   *   cen   +   2   +   (  0   if   odd   else   1  )   return   self  .  p  [  pos  ]   # checks whether substring s[l...r] is a palindrome   def   check  (  self     l     r  ):   length   =   r   -   l   +   1   longest   =   self  .  getLongest  ((  l   +   r  )   //   2     length   %   2  )   return   length    <=   longest   # returns the minimum number of characters to add at the    # front to make the given string a palindrome   def   minChar  (  s  ):   n   =   len  (  s  )   m   =   manacher  (  s  )   # scan from the end to find the longest    # palindromic prefix   for   i   in   range  (  n   -   1     -  1     -  1  ):   if   m  .  check  (  0     i  ):   return   n   -   (  i   +   1  )   return   n   -   1   if   __name__   ==   '__main__'  :   s   =   'aacecaaaa'   print  (  minChar  (  s  ))   
C#
   using     System  ;   class     GfG     {          // manacher's algorithm for finding longest       // palindromic substrings      class     manacher     {      // array to store palindrome lengths centered       // at each position      public     int  []     p  ;      // modified string with separators and sentinels      public     string     ms  ;      public     manacher  (  string     s  )     {      ms     =     '@'  ;      foreach     (  char     c     in     s  )     {      ms     +=     '#'     +     c  ;      }      ms     +=     '#$'  ;      runManacher  ();      }      // core Manacher's algorithm      void     runManacher  ()     {      int     n     =     ms  .  Length  ;      p     =     new     int  [  n  ];      int     l     =     0       r     =     0  ;      for     (  int     i     =     1  ;     i      <     n     -     1  ;     ++  i  )     {      if     (  i      <     r  )      p  [  i  ]     =     Math  .  Min  (  r     -     i       p  [  r     +     l     -     i  ]);      // expand around the current center      while     (  ms  [  i     +     1     +     p  [  i  ]]     ==     ms  [  i     -     1     -     p  [  i  ]])      p  [  i  ]  ++  ;      // update center if palindrome goes beyond       // current right boundary      if     (  i     +     p  [  i  ]     >     r  )     {      l     =     i     -     p  [  i  ];      r     =     i     +     p  [  i  ];      }      }      }      // returns the length of the longest palindrome       // centered at given position      public     int     getLongest  (  int     cen       int     odd  )     {      int     pos     =     2     *     cen     +     2     +     (  odd     ==     0     ?     1     :     0  );      return     p  [  pos  ];      }      // checks whether substring s[l...r] is a palindrome      public     bool     check  (  int     l       int     r  )     {      int     len     =     r     -     l     +     1  ;      int     longest     =     getLongest  ((  l     +     r  )     /     2       len     %     2  );      return     len      <=     longest  ;      }      }      // returns the minimum number of characters to add at the       // front to make the given string a palindrome      static     int     minChar  (  string     s  )     {      int     n     =     s  .  Length  ;      manacher     m     =     new     manacher  (  s  );      // scan from the end to find the longest       // palindromic prefix      for     (  int     i     =     n     -     1  ;     i     >=     0  ;     --  i  )     {      if     (  m  .  check  (  0       i  ))      return     n     -     (  i     +     1  );      }      return     n     -     1  ;      }      static     void     Main  ()     {      string     s     =     'aacecaaaa'  ;      Console  .  WriteLine  (  minChar  (  s  ));      }   }   
JavaScript
   // manacher's algorithm for finding longest    // palindromic substrings   class     manacher     {          // array to store palindrome lengths centered       // at each position      constructor  (  s  )     {      // modified string with separators and sentinels      this  .  ms     =     '@'  ;      for     (  let     c     of     s  )     {      this  .  ms     +=     '#'     +     c  ;      }      this  .  ms     +=     '#$'  ;      this  .  p     =     [];      this  .  runManacher  ();      }      // core Manacher's algorithm      runManacher  ()     {      const     n     =     this  .  ms  .  length  ;      this  .  p     =     new     Array  (  n  ).  fill  (  0  );      let     l     =     0       r     =     0  ;      for     (  let     i     =     1  ;     i      <     n     -     1  ;     ++  i  )     {      if     (  i      <     r  )      this  .  p  [  i  ]     =     Math  .  min  (  r     -     i       this  .  p  [  r     +     l     -     i  ]);      // expand around the current center      while     (  this  .  ms  [  i     +     1     +     this  .  p  [  i  ]]     ===     this  .  ms  [  i     -     1     -     this  .  p  [  i  ]])      this  .  p  [  i  ]  ++  ;      // update center if palindrome goes beyond       // current right boundary      if     (  i     +     this  .  p  [  i  ]     >     r  )     {      l     =     i     -     this  .  p  [  i  ];      r     =     i     +     this  .  p  [  i  ];      }      }      }      // returns the length of the longest palindrome       // centered at given position      getLongest  (  cen       odd  )     {      const     pos     =     2     *     cen     +     2     +     (  odd     ===     0     ?     1     :     0  );      return     this  .  p  [  pos  ];      }      // checks whether substring s[l...r] is a palindrome      check  (  l       r  )     {      const     len     =     r     -     l     +     1  ;      const     longest     =     this  .  getLongest  (  Math  .  floor  ((  l     +     r  )     /     2  )     len     %     2  );      return     len      <=     longest  ;      }   }   // returns the minimum number of characters to add at the    // front to make the given string a palindrome   function     minChar  (  s  )     {      const     n     =     s  .  length  ;      const     m     =     new     manacher  (  s  );      // scan from the end to find the longest       // palindromic prefix      for     (  let     i     =     n     -     1  ;     i     >=     0  ;     --  i  )     {      if     (  m  .  check  (  0       i  ))      return     n     -     (  i     +     1  );      }      return     n     -     1  ;   }   // Driver Code   const     s     =     'aacecaaaa'  ;   console  .  log  (  minChar  (  s  ));   

الإخراج
2  

تعقيد الوقت: تعمل خوارزمية O(n) manacher في وقت خطي عن طريق توسيع المتناظرات في كل مركز دون إعادة النظر في الأحرف، وتقوم حلقة التحقق من البادئة بتنفيذ عمليات O(1) لكل حرف على n من الأحرف.
المساحة المساعدة: O(n) يستخدم للسلسلة المعدلة وصفيف طول المتناظر p[] وكلاهما ينمو خطيًا مع حجم الإدخال.

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