Givet en sträng måste vi kontrollera om det är möjligt att göra denna sträng till en palindrom efter att ha tagit bort exakt ett tecken från denna.
Exempel:
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
Vi kan lösa detta problem genom att hitta läget för obalans. Vi börjar loopa in strängen genom att hålla två pekare i båda ändarna som går mot mittpositionen efter varje iteration. Denna iteration kommer att sluta när vi hittar en missmatch eftersom det är tillåtet att ta bort bara ett tecken vi har två val här
Vid oöverensstämmelse ta antingen bort tecken som pekas av vänster pekare eller ta bort tecken som pekas av höger pekare.
Vi kommer att kontrollera båda fallen, kom ihåg eftersom vi har korsat lika många steg från båda sidor. Denna mittsträng ska också vara en palindrom efter att ha tagit bort ett tecken, så vi kontrollerar två delsträngar, en genom att ta bort vänster tecken och en genom att ta bort höger tecken och om en av dem är palindrom kan vi göra en komplett strängpalindrom genom att ta bort motsvarande tecken och om båda delsträngarna inte är en palindrom att göra är det inte ett palindrom. under givna begränsningar.
Genomförande:
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>
Produktion
Not Possible
Tidskomplexitet: O(N)
Rymdkomplexitet: O(1)
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