전면 삽입에 의한 회문
GfG Practice에서 사용해 보세요. 
C++
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영어 소문자로만 구성된 문자열 s가 주어지면 최저한의 필요한 문자 수 추가됨 에 앞쪽 s를 사용하여 회문으로 만듭니다.
메모: 회문은 앞으로 읽어도 뒤로 읽어도 같은 문자열입니다.
예:
입력 : s = 'abc'
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설명 : 위 문자열 회문 앞에 'b'와 'c'를 추가하면 'cbabc'로 만들 수 있습니다.입력 : s = '아아세카아아아'
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설명 : 문자열 앞에 두 개의 a를 추가하면 위의 문자열 회문을 'aaaacecaaaa'로 만들 수 있습니다.
목차
- [순진한 접근 방식] 모든 접두사 확인 - O(n^2) 시간 및 O(1) 공간
- [예상 접근 방식 1] KMP 알고리즘의 lps 배열 사용 - O(n) 시간과 O(n) 공간
- [예상 접근 방식 2] Manacher의 알고리즘을 활용
[순진한 접근 방식] 모든 접두사 확인 - O(n^2) 시간 및 O(1) 공간
이 아이디어는 회문이기도 한 주어진 문자열에서 가장 긴 접두사를 찾아야 한다는 관찰에 기초합니다. 그런 다음 주어진 문자열 회문을 만들기 위해 추가할 최소 앞 문자는 나머지 문자가 됩니다.
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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[예상 접근 방식 1] KMP 알고리즘의 lps 배열 사용 - O(n) 시간과 O(n) 공간
중요한 관찰은 문자열의 가장 긴 회문 접두사가 그 반대 문자열의 가장 긴 회문 접미사가 된다는 것입니다.
문자열 s = 'aacecaaaa'가 주어지면 역방향 revS = 'aaaacecaa'입니다. s의 가장 긴 회문 접두사는 'aacecaa'입니다.
이를 효율적으로 찾기 위해 우리는 LPS 배열을 사용합니다. KMP 알고리즘 . 원래 문자열을 특수 문자와 그 반대 문자(s + '$' + revS)로 연결합니다.
이 결합된 문자열의 LPS 배열은 s의 회문 접두사를 나타내는 revS의 접미사와 일치하는 s의 가장 긴 접두사를 식별하는 데 도움이 됩니다.
LPS 배열의 마지막 값은 처음에 이미 회문을 형성하는 문자 수를 알려줍니다. 따라서 s를 회문으로 만들기 위해 추가해야 하는 최소 문자 수는 s.length() - lps.back()입니다.
C++ #include #include #include using namespace std ; vector < int > computeLPSArray ( string & pat ) { int n = pat . length (); vector < int > lps ( n ); // lps[0] is always 0 lps [ 0 ] = 0 ; int len = 0 ; // loop calculates lps[i] for i = 1 to M-1 int i = 1 ; while ( i < n ) { // if the characters match increment len // and set lps[i] if ( pat [ i ] == pat [ len ]) { len ++ ; lps [ i ] = len ; i ++ ; } // if there is a mismatch else { // if len is not zero update len to // the last known prefix length if ( len != 0 ) { len = lps [ len - 1 ]; } // no prefix matches set lps[i] to 0 else { lps [ i ] = 0 ; i ++ ; } } } return lps ; } // returns minimum character to be added at // front to make string palindrome int minChar ( string & s ) { int n = s . length (); string rev = s ; reverse ( rev . begin () rev . end ()); // get concatenation of string special character // and reverse string s = s + '$' + 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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[예상 접근 방식 2] Manacher의 알고리즘을 활용
C++아이디어는 사용하는 것입니다 Manacher의 알고리즘 선형 시간 내에 모든 회문 부분 문자열을 효율적으로 찾습니다.
짝수 길이와 홀수 길이의 회문을 모두 균일하게 처리하기 위해 특수 문자(#)를 삽입하여 문자열을 변환합니다.
전처리 후 원래 문자열의 끝부터 스캔하고 회문 반경 배열을 사용하여 접두사 s[0...i]가 회문인지 확인합니다. 첫 번째 인덱스 i는 가장 긴 회문 접두사를 제공하고 추가할 최소 문자로 n - (i + 1)을 반환합니다.
#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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시간 복잡도: O(n) manacher의 알고리즘은 문자를 다시 방문하지 않고 각 중심에서 회문을 확장하여 선형 시간으로 실행되며 접두사 확인 루프는 n 문자에 대해 문자당 O(1) 작업을 수행합니다.
보조 공간: 수정된 문자열과 회문 길이 배열 p[]에 사용되는 O(n)은 둘 다 입력 크기에 따라 선형적으로 증가합니다.
