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قم بإزالة حرف من سلسلة لجعله متناظرًا

بالنظر إلى سلسلة نحتاج إلى التحقق مما إذا كان من الممكن جعل هذه السلسلة متناظرة بعد إزالة حرف واحد بالضبط منها. 

التغليف في جافا

أمثلة:   

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 

يمكننا حل هذه المشكلة من خلال إيجاد موضع عدم التطابق. نبدأ في التكرار في السلسلة عن طريق الاحتفاظ بمؤشرين في كلا الطرفين يتجهان نحو الموضع الأوسط بعد كل تكرار، سيتوقف هذا التكرار عندما نجد عدم تطابق حيث يُسمح بإزالة حرف واحد فقط لدينا خياران هنا



في حالة عدم التطابق، قم بإزالة الحرف الذي يشير إليه المؤشر الأيسر أو قم بإزالة الحرف الذي يشير إليه المؤشر الأيمن.

سوف نتحقق من كلتا الحالتين، تذكر أننا قد اجتزنا عددًا متساويًا من الخطوات من كلا الجانبين، ويجب أن تكون هذه السلسلة الوسطى أيضًا متناظرة بعد إزالة حرف واحد، لذلك نتحقق من سلسلتين فرعيتين واحدة عن طريق إزالة الحرف الأيسر وواحدة عن طريق إزالة الحرف الأيمن وإذا كانت إحداهما متناظرة، فيمكننا إنشاء سلسلة متناظرة كاملة عن طريق إزالة الحرف المقابل، وإذا لم تكن كلا السلسلتين الفرعيتين متناظرتين، فمن غير الممكن جعل السلسلة الكاملة متناظرة تحت قيد معين. 

تطبيق:

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> 

الإخراج
Not Possible 

تعقيد الوقت : O(N)
تعقيد الفضاء: O(1)

 

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