Palindrom s sprednjim vstavljanjem
Preizkusite na GfG Practice 
C++
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Podan je niz s, ki je sestavljen samo iz malih angleških črk, poiščite najmanj število znakov, ki morajo biti dodano do spredaj od s, da postane palindrom.
Opomba: Palindrom je niz, ki se enako bere naprej in nazaj.
Primeri:
Vnos : s = 'abc'
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Pojasnilo : Zgornji palindrom niza lahko naredimo kot 'cbabc' tako, da spredaj dodamo 'b' in 'c'.Vnos : s = 'aacecaaaa'
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Razlaga : Zgornji palindrom niza lahko naredimo kot 'aaaacecaaaa' tako, da dodamo dva a pred niz.
Kazalo vsebine
- [Naivni pristop] Preverjanje vseh predpon - O(n^2) Čas in O(1) Prostor
- [Pričakovani pristop 1] Uporaba polja lps algoritma KMP – O(n) časa in O(n) prostora
- [Pričakovani pristop 2] Uporaba Manacherjevega algoritma
[Naivni pristop] Preverjanje vseh predpon - O(n^2) Čas in O(1) Prostor
Ideja temelji na ugotovitvi, da moramo najti najdaljšo predpono iz danega niza, ki je hkrati palindrom. Nato bodo preostali znaki minimalni sprednji znaki, ki jih je treba dodati, da sestavi dani palindrom niza.
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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[Pričakovani pristop 1] Uporaba polja lps algoritma KMP – O(n) časa in O(n) prostora
Ključna ugotovitev je, da najdaljša palindromna predpona niza postane najdaljša palindromna pripona njegovega obrata.
Podan je niz s = 'aacecaaaa', njegov obratni revS = 'aaaacecaa'. Najdaljša palindromna predpona s je 'aacecaa'.
Za učinkovito iskanje tega uporabljamo niz LPS iz algoritem KMP . Izvirni niz združimo s posebnim znakom in njegovo obratno stran: s + '$' + revS.
Matrika LPS za ta kombinirani niz pomaga identificirati najdaljšo predpono s, ki se ujema s pripono revS, ki prav tako predstavlja palindromsko predpono s.
Zadnja vrednost matrike LPS nam pove, koliko znakov že na začetku tvori palindrom. Tako je najmanjše število znakov, ki jih je treba dodati, da s postane palindrom, s.length() - lps.back().
C++ #include #include #include using namespace std ; vector < int > computeLPSArray ( string & pat ) { int n = pat . length (); vector < int > lps ( n ); // lps[0] is always 0 lps [ 0 ] = 0 ; int len = 0 ; // loop calculates lps[i] for i = 1 to M-1 int i = 1 ; while ( i < n ) { // if the characters match increment len // and set lps[i] if ( pat [ i ] == pat [ len ]) { len ++ ; lps [ i ] = len ; i ++ ; } // if there is a mismatch else { // if len is not zero update len to // the last known prefix length if ( len != 0 ) { len = lps [ len - 1 ]; } // no prefix matches set lps[i] to 0 else { lps [ i ] = 0 ; i ++ ; } } } return lps ; } // returns minimum character to be added at // front to make string palindrome int minChar ( string & s ) { int n = s . length (); string rev = s ; reverse ( rev . begin () rev . end ()); // get concatenation of string special character // and reverse string s = s + '$' + 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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[Pričakovani pristop 2] Uporaba Manacherjevega algoritma
C++Ideja je uporaba Manacherjev algoritem za učinkovito iskanje vseh palindromskih podnizov v linearnem času.
Niz preoblikujemo tako, da vstavimo posebne znake (#), da enakomerno obravnavamo palindrome sode in lihe dolžine.
Po predobdelavi skeniramo od konca izvirnega niza in uporabimo matriko polmerov palindroma, da preverimo, ali je predpona s[0...i] palindrom. Prvi tak indeks i nam daje najdaljšo palindromsko predpono in vrnemo n - (i + 1) kot najmanjše število znakov, ki jih je treba dodati.
#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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Časovna zapletenost: O(n) manacherjev algoritem teče v linearnem času z razširitvijo palindromov v vsakem središču brez ponovnega pregleda znakov in zanka za preverjanje predpone izvede O(1) operacij na znak v n znakih.
Pomožni prostor: O(n), ki se uporablja za spremenjeni niz in niz dolžin palindroma p[], ki oba rasteta linearno z velikostjo vnosa.
