Palindrom durch Fronteinfügung
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C++
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Gegeben sei eine Zeichenfolge, die nur aus englischen Kleinbuchstaben besteht, finden Sie die Minimum Anzahl der Zeichen, die sein müssen hinzugefügt zum Front von s, um daraus ein Palindrom zu machen.
Notiz: Ein Palindrom ist eine Zeichenfolge, die vorwärts und rückwärts dasselbe liest.
Beispiele:
Eingang : s = 'abc'
Ausgabe : 2
Erläuterung : Wir können das obige String-Palindrom als „cbabc“ erstellen, indem wir „b“ und „c“ vorne hinzufügen.Eingang : s = 'aacecaaaa'
Ausgabe : 2
Erläuterung : Wir können das obige String-Palindrom als „aaaacecaaaa“ erstellen, indem wir zwei a vor dem String hinzufügen.
Inhaltsverzeichnis
- [Naiver Ansatz] Überprüfung aller Präfixe – O(n^2) Zeit und O(1) Raum
- [Erwarteter Ansatz 1] Verwendung des lps-Arrays des KMP-Algorithmus – O(n) Zeit und O(n) Raum
- [Erwarteter Ansatz 2] Verwendung des Manacher-Algorithmus
[Naiver Ansatz] Überprüfung aller Präfixe – O(n^2) Zeit und O(1) Raum
Die Idee basiert auf der Beobachtung, dass wir das längste Präfix aus einer bestimmten Zeichenfolge finden müssen, die auch ein Palindrom ist. Dann sind die verbleibenden Zeichen die minimalen vorderen Zeichen, die hinzugefügt werden müssen, um eine gegebene Zeichenfolge zu einem Palindrom zu machen.
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 ));
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2
[Erwarteter Ansatz 1] Verwendung des lps-Arrays des KMP-Algorithmus – O(n) Zeit und O(n) Raum
Die wichtigste Beobachtung ist, dass das längste palindromische Präfix einer Zeichenfolge zum längsten palindromischen Suffix ihrer Umkehrung wird.
Gegeben sei ein String s = 'aacecaaaa', dessen Umkehrung revS = 'aaaacecaa' sei. Das längste palindromische Präfix von s ist „aacecaa“.
Um dies effizient zu finden, verwenden wir das LPS-Array von KMP-Algorithmus . Wir verketten den ursprünglichen String mit einem Sonderzeichen und dessen Umkehrung: s + '$' + revS.
Das LPS-Array für diese kombinierte Zeichenfolge hilft dabei, das längste Präfix von s zu identifizieren, das mit einem Suffix von revS übereinstimmt, das auch das palindromische Präfix von s darstellt.
Der letzte Wert des LPS-Arrays sagt uns, wie viele Zeichen bereits zu Beginn ein Palindrom bilden. Daher beträgt die Mindestanzahl an Zeichen, die hinzugefügt werden müssen, um s zu einem Palindrom zu machen, s.length() - lps.back().
C++ #include #include #include using namespace std ; vector < int > computeLPSArray ( string & pat ) { int n = pat . length (); vector < int > lps ( n ); // lps[0] is always 0 lps [ 0 ] = 0 ; int len = 0 ; // loop calculates lps[i] for i = 1 to M-1 int i = 1 ; while ( i < n ) { // if the characters match increment len // and set lps[i] if ( pat [ i ] == pat [ len ]) { len ++ ; lps [ i ] = len ; i ++ ; } // if there is a mismatch else { // if len is not zero update len to // the last known prefix length if ( len != 0 ) { len = lps [ len - 1 ]; } // no prefix matches set lps[i] to 0 else { lps [ i ] = 0 ; i ++ ; } } } return lps ; } // returns minimum character to be added at // front to make string palindrome int minChar ( string & s ) { int n = s . length (); string rev = s ; reverse ( rev . begin () rev . end ()); // get concatenation of string special character // and reverse string s = s + '$' + 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 ));
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[Erwarteter Ansatz 2] Verwendung des Manacher-Algorithmus
C++Die Idee ist zu verwenden Manachers Algorithmus um alle palindromischen Teilzeichenfolgen effizient in linearer Zeit zu finden.
Wir transformieren die Zeichenfolge, indem wir Sonderzeichen (#) einfügen, um Palindrome gerader und ungerader Länge einheitlich zu behandeln.
Nach der Vorverarbeitung scannen wir vom Ende des ursprünglichen Strings und verwenden das Palindrom-Radius-Array, um zu prüfen, ob das Präfix s[0...i] ein Palindrom ist. Der erste dieser Indexe i gibt uns das längste palindromische Präfix und wir geben n - (i + 1) als minimal hinzuzufügende Zeichen zurück.
#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 ));
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Zeitkomplexität: Der O(n)-Manacher-Algorithmus läuft in linearer Zeit, indem er Palindrome in jedem Zentrum erweitert, ohne Zeichen erneut aufzurufen, und die Präfixprüfschleife führt O(1)-Operationen pro Zeichen über n Zeichen durch.
Hilfsraum: O(n) wird für die geänderte Zeichenfolge und das Palindrom-Längenarray p[] verwendet, die beide linear mit der Eingabegröße wachsen.
