Encontre o palíndromo mais longo formado removendo ou embaralhando caracteres da string
Dada uma string, encontre o palíndromo mais longo que pode ser construído removendo ou embaralhando caracteres da string. Retorne apenas um palíndromo se houver várias strings de palíndromo de maior comprimento.
Exemplos:
Input: abc Output: a OR b OR c Input: aabbcc Output: abccba OR baccab OR cbaabc OR any other palindromic string of length 6. Input: abbaccd Output: abcdcba OR ... Input: aba Output: aba
Podemos dividir qualquer corda palindrômica em três partes - início, meio e fim. Para string palindrômica de comprimento ímpar, digamos 2n + 1 'beg' consiste nos primeiros n caracteres da string 'mid' consistirá em apenas 1 caractere, ou seja, (n + 1) o caractere e 'end' consistirá nos últimos n caracteres da string palindrômica. Para sequência palindrômica de comprimento par 2n, 'mid' sempre estará vazio. Deve-se notar que 'end' será o inverso de 'beg' para que a string seja palíndromo.
A ideia é usar a observação acima em nossa solução. Como o embaralhamento de caracteres é permitido, a ordem dos caracteres não importa na string de entrada. Primeiro obtemos a frequência de cada caractere na string de entrada. Então, todos os caracteres com ocorrência par (digamos 2n) na string de entrada farão parte da string de saída, pois podemos facilmente colocar n caracteres na string 'beg' e os outros n caracteres na string 'end' (preservando a ordem palindrômica). Para caracteres de ocorrência ímpar (digamos 2n + 1), preenchemos 'mid' com um de todos esses caracteres. e os 2n caracteres restantes são divididos ao meio e adicionados no início e no final.
Abaixo está a implementação da ideia acima
C++ // C++ program to find the longest palindrome by removing // or shuffling characters from the given string #include using namespace std ; // Function to find the longest palindrome by removing // or shuffling characters from the given string string findLongestPalindrome ( string str ) { // to stores freq of characters in a string int count [ 256 ] = { 0 }; // find freq of characters in the input string for ( int i = 0 ; i < str . size (); i ++ ) count [ str [ i ]] ++ ; // Any palindromic string consists of three parts // beg + mid + end string beg = '' mid = '' end = '' ; // solution assumes only lowercase characters are // present in string. We can easily extend this // to consider any set of characters for ( char ch = 'a' ; ch <= 'z' ; ch ++ ) { // if the current character freq is odd if ( count [ ch ] & 1 ) { // mid will contain only 1 character. It // will be overridden with next character // with odd freq mid = ch ; // decrement the character freq to make // it even and consider current character // again count [ ch -- ] -- ; } // if the current character freq is even else { // If count is n(an even number) push // n/2 characters to beg string and rest // n/2 characters will form part of end // string for ( int i = 0 ; i < count [ ch ] / 2 ; i ++ ) beg . push_back ( ch ); } } // end will be reverse of beg end = beg ; reverse ( end . begin () end . end ()); // return palindrome string return beg + mid + end ; } // Driver code int main () { string str = 'abbaccd' ; cout < < findLongestPalindrome ( str ); return 0 ; }
Java // Java program to find the longest palindrome by removing // or shuffling characters from the given string class GFG { // Function to find the longest palindrome by removing // or shuffling characters from the given string static String findLongestPalindrome ( String str ) { // to stores freq of characters in a string int count [] = new int [ 256 ] ; // find freq of characters in the input string for ( int i = 0 ; i < str . length (); i ++ ) { count [ str . charAt ( i ) ]++ ; } // Any palindromic string consists of three parts // beg + mid + end String beg = '' mid = '' end = '' ; // solution assumes only lowercase characters are // present in string. We can easily extend this // to consider any set of characters for ( char ch = 'a' ; ch <= 'z' ; ch ++ ) { // if the current character freq is odd if ( count [ ch ] % 2 == 1 ) { // mid will contain only 1 character. It // will be overridden with next character // with odd freq mid = String . valueOf ( ch ); // decrement the character freq to make // it even and consider current character // again count [ ch --]-- ; } // if the current character freq is even else { // If count is n(an even number) push // n/2 characters to beg string and rest // n/2 characters will form part of end // string for ( int i = 0 ; i < count [ ch ] / 2 ; i ++ ) { beg += ch ; } } } // end will be reverse of beg end = beg ; end = reverse ( end ); // return palindrome string return beg + mid + end ; } static String reverse ( String str ) { // convert String to character array // by using toCharArray String ans = '' ; char [] try1 = str . toCharArray (); for ( int i = try1 . length - 1 ; i >= 0 ; i -- ) { ans += try1 [ i ] ; } return ans ; } // Driver code public static void main ( String [] args ) { String str = 'abbaccd' ; System . out . println ( findLongestPalindrome ( str )); } } // This code is contributed by PrinciRaj1992
Python3 # Python3 program to find the longest palindrome by removing # or shuffling characters from the given string # Function to find the longest palindrome by removing # or shuffling characters from the given string def findLongestPalindrome ( strr ): # to stores freq of characters in a string count = [ 0 ] * 256 # find freq of characters in the input string for i in range ( len ( strr )): count [ ord ( strr [ i ])] += 1 # Any palindromic consists of three parts # beg + mid + end beg = '' mid = '' end = '' # solution assumes only lowercase characters are # present in string. We can easily extend this # to consider any set of characters ch = ord ( 'a' ) while ch <= ord ( 'z' ): # if the current character freq is odd if ( count [ ch ] & 1 ): # mid will contain only 1 character. It # will be overridden with next character # with odd freq mid = ch # decrement the character freq to make # it even and consider current character # again count [ ch ] -= 1 ch -= 1 # if the current character freq is even else : # If count is n(an even number) push # n/2 characters to beg and rest # n/2 characters will form part of end # string for i in range ( count [ ch ] // 2 ): beg += chr ( ch ) ch += 1 # end will be reverse of beg end = beg end = end [:: - 1 ] # return palindrome string return beg + chr ( mid ) + end # Driver code strr = 'abbaccd' print ( findLongestPalindrome ( strr )) # This code is contributed by mohit kumar 29
C# // C# program to find the longest // palindrome by removing or // shuffling characters from // the given string using System ; class GFG { // Function to find the longest // palindrome by removing or // shuffling characters from // the given string static String findLongestPalindrome ( String str ) { // to stores freq of characters in a string int [] count = new int [ 256 ]; // find freq of characters // in the input string for ( int i = 0 ; i < str . Length ; i ++ ) { count [ str [ i ]] ++ ; } // Any palindromic string consists of // three parts beg + mid + end String beg = '' mid = '' end = '' ; // solution assumes only lowercase // characters are present in string. // We can easily extend this to // consider any set of characters for ( char ch = 'a' ; ch <= 'z' ; ch ++ ) { // if the current character freq is odd if ( count [ ch ] % 2 == 1 ) { // mid will contain only 1 character. // It will be overridden with next // character with odd freq mid = String . Join ( '' ch ); // decrement the character freq to make // it even and consider current // character again count [ ch -- ] -- ; } // if the current character freq is even else { // If count is n(an even number) push // n/2 characters to beg string and rest // n/2 characters will form part of end // string for ( int i = 0 ; i < count [ ch ] / 2 ; i ++ ) { beg += ch ; } } } // end will be reverse of beg end = beg ; end = reverse ( end ); // return palindrome string return beg + mid + end ; } static String reverse ( String str ) { // convert String to character array // by using toCharArray String ans = '' ; char [] try1 = str . ToCharArray (); for ( int i = try1 . Length - 1 ; i >= 0 ; i -- ) { ans += try1 [ i ]; } return ans ; } // Driver code public static void Main () { String str = 'abbaccd' ; Console . WriteLine ( findLongestPalindrome ( str )); } } // This code is contributed by 29AjayKumar
JavaScript < script > // Javascript program to find the // longest palindrome by removing // or shuffling characters from // the given string // Function to find the longest // palindrome by removing // or shuffling characters from // the given string function findLongestPalindrome ( str ) { // to stores freq of characters // in a string let count = new Array ( 256 ); for ( let i = 0 ; i < 256 ; i ++ ) { count [ i ] = 0 ; } // find freq of characters in // the input string for ( let i = 0 ; i < str . length ; i ++ ) { count [ str [ i ]. charCodeAt ( 0 )] ++ ; } // Any palindromic string consists // of three parts // beg + mid + end let beg = '' mid = '' end = '' ; // solution assumes only // lowercase characters are // present in string. // We can easily extend this // to consider any set of characters for ( let ch = 'a' . charCodeAt ( 0 ); ch <= 'z' . charCodeAt ( 0 ); ch ++ ) { // if the current character freq is odd if ( count [ ch ] % 2 == 1 ) { // mid will contain only 1 character. It // will be overridden with next character // with odd freq mid = String . fromCharCode ( ch ); // decrement the character freq to make // it even and consider current character // again count [ ch -- ] -- ; } // if the current character freq is even else { // If count is n(an even number) push // n/2 characters to beg string and rest // n/2 characters will form part of end // string for ( let i = 0 ; i < count [ ch ] / 2 ; i ++ ) { beg += String . fromCharCode ( ch ); } } } // end will be reverse of beg end = beg ; end = reverse ( end ); // return palindrome string return beg + mid + end ; } function reverse ( str ) { // convert String to character array // by using toCharArray let ans = '' ; let try1 = str . split ( '' ); for ( let i = try1 . length - 1 ; i >= 0 ; i -- ) { ans += try1 [ i ]; } return ans ; } // Driver code let str = 'abbaccd' ; document . write ( findLongestPalindrome ( str )); // This code is contributed by unknown2108 < /script>
Saída
abcdcba
Complexidade de tempo da solução acima é O(n) onde n é o comprimento da string. Como o número de caracteres do alfabeto é constante, eles não contribuem para a análise assintótica.
Espaço auxiliar usado pelo programa é M onde M é o número de caracteres ASCII.
