Palindromas naudojant priekinį įdėjimą
Išbandykite GfG praktikoje 
C++
Išvestis
Išvestis
Išvestis
Atsižvelgdami į eilutę s, kurią sudaro tik mažosios angliškos raidės, raskite minimumas simbolių skaičius, kuris turi būti pridėta prie priekyje s, kad jis taptų palindromu.
Pastaba: Palindromas yra eilutė, kuri skaito tą pačią pirmyn ir atgal.
Pavyzdžiai:
Įvestis : s = 'abc'
Išvestis : 2
Paaiškinimas : Virš eilutės esantį palindromą galime padaryti kaip „cbabc“, priekyje pridėdami „b“ ir „c“.Įvestis : s = 'aacecaaaa'
Išvestis : 2
Paaiškinimas : Virš eilutės esantį palindromą galime padaryti kaip „aaaacecaaaa“, prie eilutės pridėję du a.
Turinio lentelė
- [Naivus požiūris] Tikrinami visi priešdėliai – O(n^2) laikas ir O(1) erdvė
- [Numatomas 1 metodas] Naudojant KMP algoritmo lps masyvą – O(n) laikas ir O(n) erdvė
- [Numatomas 2 metodas] Naudojant Manacherio algoritmą
[Naivus požiūris] Tikrinami visi priešdėliai – O(n^2) laikas ir O(1) erdvė
Idėja pagrįsta pastebėjimu, kad turime rasti ilgiausią priešdėlį iš nurodytos eilutės, kuri taip pat yra palindromas. Tada minimalūs priekiniai simboliai, kuriuos reikia pridėti, kad būtų sukurtas tam tikros eilutės palindromas, bus likę simboliai.
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 ));
Išvestis
2
[Numatomas 1 metodas] Naudojant KMP algoritmo lps masyvą – O(n) laikas ir O(n) erdvė
Pagrindinis pastebėjimas yra tas, kad ilgiausias palindrominis stygos priešdėlis tampa ilgiausia palindromine jos reverso priesaga.
Jei eilutė s = 'aacecaaaa', jos atvirkštinė revS = 'aaaacecaa'. Ilgiausias palindrominis s priešdėlis yra „aacecaa“.
Norėdami tai efektyviai rasti, naudojame LPS masyvą iš KMP algoritmas . Pradinę eilutę sujungiame su specialiu simboliu ir jo atvirkštine dalimi: s + '$' + revS.
Šios kombinuotos eilutės LPS masyvas padeda nustatyti ilgiausią s priešdėlį, atitinkantį revS galūnę, kuri taip pat reiškia palindrominį s priešdėlį.
Paskutinė LPS masyvo reikšmė nurodo, kiek simbolių jau pradžioje sudaro palindromą. Taigi minimalus simbolių skaičius, kurį reikia pridėti, kad būtų palindromas, yra s.length() - lps.back().
C++ #include #include #include using namespace std ; vector < int > computeLPSArray ( string & pat ) { int n = pat . length (); vector < int > lps ( n ); // lps[0] is always 0 lps [ 0 ] = 0 ; int len = 0 ; // loop calculates lps[i] for i = 1 to M-1 int i = 1 ; while ( i < n ) { // if the characters match increment len // and set lps[i] if ( pat [ i ] == pat [ len ]) { len ++ ; lps [ i ] = len ; i ++ ; } // if there is a mismatch else { // if len is not zero update len to // the last known prefix length if ( len != 0 ) { len = lps [ len - 1 ]; } // no prefix matches set lps[i] to 0 else { lps [ i ] = 0 ; i ++ ; } } } return lps ; } // returns minimum character to be added at // front to make string palindrome int minChar ( string & s ) { int n = s . length (); string rev = s ; reverse ( rev . begin () rev . end ()); // get concatenation of string special character // and reverse string s = s + '$' + 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 ));
Išvestis
2
[Numatomas 2 metodas] Naudojant Manacherio algoritmą
C++Idėja yra naudoti Manacherio algoritmas efektyviai rasti visas palindromines eilutes tiesiniu laiku.
Mes transformuojame eilutę įterpdami specialiuosius simbolius (#), kad vienodai tvarkytume lyginio ir nelyginio ilgio palindromus.
Po išankstinio apdorojimo nuskaitome nuo pradinės eilutės pabaigos ir naudojame palindromo spindulio masyvą, kad patikrintume, ar priešdėlis s[0...i] yra palindromas. Pirmasis toks indeksas i suteikia mums ilgiausią palindrominį priešdėlį, o mes grąžiname n - (i + 1) kaip minimalius pridedamus simbolius.
#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 ));
Išvestis
2
Laiko sudėtingumas: O(n) vadovo algoritmas veikia tiesiniu laiku, kiekviename centre išplečiant palindromus, neperžiūrint simbolių, o priešdėlio tikrinimo kilpa atlieka O(1) operacijas kiekvienam simboliui per n simbolių.
Pagalbinė erdvė: O(n), naudojamas modifikuotai eilutei ir palindromo ilgio masyvei p[], kurie abu auga tiesiškai atsižvelgiant į įvesties dydį.
