Supprimer un caractère d'une chaîne pour en faire un palindrome
Étant donné une chaîne, nous devons vérifier s'il est possible de faire de cette chaîne un palindrome après en avoir supprimé exactement un caractère.
Exemples :
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
Nous pouvons résoudre ce problème en trouvant la position du décalage. Nous commençons à boucler la chaîne en gardant deux pointeurs aux deux extrémités qui se déplacent vers la position médiane après chaque itération. Cette itération s'arrêtera lorsque nous trouverons une incompatibilité car il est autorisé de supprimer un seul caractère. Nous avons deux choix ici.
En cas de non-concordance, supprimez le caractère pointé par le pointeur gauche ou supprimez le caractère pointé par le pointeur droit.
Nous vérifierons les deux cas, rappelez-vous que nous avons parcouru un nombre égal d'étapes des deux côtés, cette chaîne médiane devrait également être un palindrome après avoir supprimé un caractère. Nous vérifions donc deux sous-chaînes, une en supprimant le caractère de gauche et une en supprimant le caractère de droite.
Mise en œuvre:
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>
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Complexité temporelle : O(N)
Complexité spatiale : O(1)
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