Le plus petit palindrome après remplacement
Étant donné une chaîne contenant des caractères alphabétiques minuscules et un point de caractère spécial (.). Nous devons remplacer tous les points par un caractère alphabétique de telle sorte que la chaîne résultante devienne un palindrome. En cas de nombreux remplacements possibles, nous devons choisir la chaîne palindrome qui est lexicographiquement la plus petite. S'il n'est pas possible de convertir une chaîne en palindrome après tous les remplacements possibles, affichez Not possible.
Exemples :
Input : str = ab..e.c.a Output : abcaeacba The smallest palindrome which can be made after replacement is 'abcaeacba' We replaced first dot with 'c' second dot with 'a' third dot with 'a' and fourth dot with 'b' Input : str = ab..e.c.b Output : Not Possible It is not possible to convert above string into palindrome
Nous pouvons résoudre ce problème comme suit. Comme la chaîne résultante doit être un palindrome, nous pouvons vérifier une paire de caractères autres que des points au démarrage s'ils ne correspondent pas, le retour direct n'est alors pas possible car nous pouvons placer un nouveau caractère à la position des points uniquement et nulle part ailleurs.
Après cela, nous parcourons les caractères de la chaîne si le caractère actuel est un point, puis nous vérifions son caractère apparié (caractère à la (n – i -1)ème position) si ce caractère est également un point, nous pouvons remplacer les deux caractères par « a » car « a » est le plus petit alphabet minuscule qui garantira la plus petite chaîne lexicographique à la fin, le remplacement des deux par n'importe quel autre caractère entraînera une chaîne palindromique lexicographiquement plus grande. Dans les autres cas, si le caractère apparié n'est pas un point, alors pour créer un palindrome de chaîne, nous devons remplacer le caractère actuel par son caractère apparié.
So in short If both 'i' and 'n- i- 1' are dot replace them by ‘a’ If one of them is a dot character replace that by other non-dot character
La procédure ci-dessus nous donne la plus petite chaîne de palindrome lexicographiquement.
Mise en œuvre:
C++ // C++ program to get lexicographically smallest // palindrome string #include using namespace std ; // Utility method to check str is possible palindrome // after ignoring . bool isPossiblePalindrome ( string str ) { int n = str . length (); for ( int i = 0 ; i < n / 2 ; i ++ ) { /* If both left and right character are not dot and they are not equal also then it is not possible to make this string a palindrome */ if ( str [ i ] != '.' && str [ n - i -1 ] != '.' && str [ i ] != str [ n - i -1 ]) return false ; } return true ; } // Returns lexicographically smallest palindrom // string if possible string smallestPalindrome ( string str ) { if ( ! isPossiblePalindrome ( str )) return 'Not Possible' ; int n = str . length (); // loop through character of string for ( int i = 0 ; i < n ; i ++ ) { if ( str [ i ] == '.' ) { // if one of character is dot replace dot // with other character if ( str [ n - i - 1 ] != '.' ) str [ i ] = str [ n - i - 1 ]; // if both character are dot then replace // them with smallest character 'a' else str [ i ] = str [ n - i - 1 ] = 'a' ; } } // return the result return str ; } // Driver code to test above methods int main () { string str = 'ab..e.c.a' ; cout < < smallestPalindrome ( str ) < < endl ; return 0 ; }
Java // Java program to get lexicographically // smallest palindrome string class GFG { // Utility method to check str is // possible palindrome after ignoring static boolean isPossiblePalindrome ( char str [] ) { int n = str . length ; for ( int i = 0 ; i < n / 2 ; i ++ ) { /* If both left and right character are not dot and they are not equal also then it is not possible to make this string a palindrome */ if ( str [ i ] != '.' && str [ n - i - 1 ] != '.' && str [ i ] != str [ n - i - 1 ] ) return false ; } return true ; } // Returns lexicographically smallest // palindrome string if possible static void smallestPalindrome ( char str [] ) { if ( ! isPossiblePalindrome ( str )) System . out . println ( 'Not Possible' ); int n = str . length ; // loop through character of string for ( int i = 0 ; i < n ; i ++ ) { if ( str [ i ] == '.' ) { // if one of character is dot // replace dot with other character if ( str [ n - i - 1 ] != '.' ) str [ i ] = str [ n - i - 1 ] ; // if both character are dot // then replace them with // smallest character 'a' else str [ i ] = str [ n - i - 1 ] = 'a' ; } } // return the result for ( int i = 0 ; i < n ; i ++ ) System . out . print ( str [ i ] + '' ); } // Driver code public static void main ( String [] args ) { String str = 'ab..e.c.a' ; char [] s = str . toCharArray (); smallestPalindrome ( s ); } } // This code is contributed // by ChitraNayal
Python 3 # Python 3 program to get lexicographically # smallest palindrome string # Utility method to check str is # possible palindrome after ignoring def isPossiblePalindrome ( str ): n = len ( str ) for i in range ( n // 2 ): # If both left and right character # are not dot and they are not # equal also then it is not possible # to make this string a palindrome if ( str [ i ] != '.' and str [ n - i - 1 ] != '.' and str [ i ] != str [ n - i - 1 ]): return False return True # Returns lexicographically smallest # palindrome string if possible def smallestPalindrome ( str ): if ( not isPossiblePalindrome ( str )): return 'Not Possible' n = len ( str ) str = list ( str ) # loop through character of string for i in range ( n ): if ( str [ i ] == '.' ): # if one of character is dot # replace dot with other character if ( str [ n - i - 1 ] != '.' ): str [ i ] = str [ n - i - 1 ] # if both character are dot # then replace them with # smallest character 'a' else : str [ i ] = str [ n - i - 1 ] = 'a' # return the result return str # Driver code if __name__ == '__main__' : str = 'ab..e.c.a' print ( '' . join ( smallestPalindrome ( str ))) # This code is contributed by ChitraNayal
C# // C# program to get lexicographically // smallest palindrome string using System ; public class GFG { // Utility method to check str is // possible palindrome after ignoring static bool isPossiblePalindrome ( char [] str ) { int n = str . Length ; for ( int i = 0 ; i < n / 2 ; i ++ ) { /* If both left and right character are not dot and they are not equal also then it is not possible to make this string a palindrome */ if ( str [ i ] != '.' && str [ n - i - 1 ] != '.' && str [ i ] != str [ n - i - 1 ]) return false ; } return true ; } // Returns lexicographically smallest // palindrome string if possible static void smallestPalindrome ( char [] str ) { if ( ! isPossiblePalindrome ( str )) Console . WriteLine ( 'Not Possible' ); int n = str . Length ; // loop through character of string for ( int i = 0 ; i < n ; i ++ ) { if ( str [ i ] == '.' ) { // if one of character is dot // replace dot with other character if ( str [ n - i - 1 ] != '.' ) str [ i ] = str [ n - i - 1 ]; // if both character are dot // then replace them with // smallest character 'a' else str [ i ] = str [ n - i - 1 ] = 'a' ; } } // return the result for ( int i = 0 ; i < n ; i ++ ) Console . Write ( str [ i ] + '' ); } // Driver code public static void Main () { String str = 'ab..e.c.a' ; char [] s = str . ToCharArray (); smallestPalindrome ( s ); } } // This code is contributed by PrinciRaj1992
PHP // PHP program to get lexicographically // smallest palindrome string // Utility method to check str is // possible palindrome after ignoring function isPossiblePalindrome ( $str ) { $n = strlen ( $str ); for ( $i = 0 ; $i < $n / 2 ; $i ++ ) { /* If both left and right character are not dot and they are not equal also then it is not possible to make this string a palindrome */ if ( $str [ $i ] != '.' && $str [ $n - $i - 1 ] != '.' && $str [ $i ] != $str [ $n - $i - 1 ]) return false ; } return true ; } // Returns lexicographically smallest // palindrome string if possible function smallestPalindrome ( $str ) { if ( ! isPossiblePalindrome ( $str )) return 'Not Possible' ; $n = strlen ( $str ); // loop through character of string for ( $i = 0 ; $i < $n ; $i ++ ) { if ( $str [ $i ] == '.' ) { // if one of character is dot // replace dot with other character if ( $str [ $n - $i - 1 ] != '.' ) $str [ $i ] = $str [ $n - $i - 1 ]; // if both character are dot // then replace them with // smallest character 'a' else $str [ $i ] = $str [ $n - $i - 1 ] = 'a' ; } } // return the result return $str ; } // Driver code $str = 'ab..e.c.a' ; echo smallestPalindrome ( $str ); // This code is contributed // by ChitraNayal ?>
JavaScript < script > // Javascript program to get lexicographically // smallest palindrome string // Utility method to check str is // possible palindrome after ignoring function isPossiblePalindrome ( str ) { let n = str . length ; for ( let i = 0 ; i < Math . floor ( n / 2 ); i ++ ) { /* If both left and right character are not dot and they are not equal also then it is not possible to make this string a palindrome */ if ( str [ i ] != '.' && str [ n - i - 1 ] != '.' && str [ i ] != str [ n - i - 1 ]) return false ; } return true ; } // Returns lexicographically smallest // palindrome string if possible function smallestPalindrome ( str ) { if ( ! isPossiblePalindrome ( str )) document . write ( 'Not Possible' ); let n = str . length ; // loop through character of string for ( let i = 0 ; i < n ; i ++ ) { if ( str [ i ] == '.' ) { // if one of character is dot // replace dot with other character if ( str [ n - i - 1 ] != '.' ) str [ i ] = str [ n - i - 1 ]; // if both character are dot // then replace them with // smallest character 'a' else str [ i ] = str [ n - i - 1 ] = 'a' ; } } // return the result for ( let i = 0 ; i < n ; i ++ ) document . write ( str [ i ] + '' ); } // Driver code let str = 'ab..e.c.a' ; let s = str . split ( '' ); smallestPalindrome ( s ); // This code is contributed by rag2127 < /script>
Sortir
abcaeacba
Complexité temporelle : O(n) où n est la longueur de la chaîne.
Complexité de l'espace auxiliaire : O(1)
