Hitta den längsta palindromen som bildas genom att ta bort eller blanda chars från strängen
Givet en sträng hitta den längsta palindromen som kan konstrueras genom att ta bort eller blanda tecken från strängen. Returnera endast en palindrom om det finns flera palindromsträngar med längst längd.
Exempel:
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
Vi kan dela vilken palindromisk sträng som helst i tre delar - börja mitten och slutet. För palindromisk sträng med udda längd säg att 2n + 1 'beg' består av de första n tecknen i strängen 'mid' kommer att bestå av endast 1 tecken, dvs (n + 1):e tecknet och 'end' består av de sista n tecknen i den palindromiska strängen. För palindromisk sträng med jämn längd 2n kommer 'mid' alltid att vara tom. Det bör noteras att 'slut' kommer att vara omvänt till 'tigga' för att strängen ska vara palindrom.
Tanken är att använda ovanstående observation i vår lösning. Eftersom blandning av tecken är tillåten spelar ordningen av tecken ingen roll i inmatningssträngen. Vi får först frekvensen för varje tecken i inmatningssträngen. Då kommer alla tecken som har jämn förekomst (säg 2n) i inmatningssträngen att vara en del av utdatasträngen eftersom vi enkelt kan placera n tecken i 'beg'-strängen och de andra n tecknen i 'slut'-strängen (genom att bevara den palindromiska ordningen). För tecken med udda förekomst (säg 2n + 1) fyller vi 'mid' med ett av alla sådana tecken. och återstående 2n tecken delas i halvor och läggs till i början och slutet.
Nedan är implementeringen av ovanstående idé
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>
Produktion
abcdcba
Tidskomplexitet av ovanstående lösning är O(n) där n är längden på strängen. Eftersom antalet tecken i alfabetet är konstant bidrar de inte till asymptotisk analys.
Hjälputrymme som används av programmet är M där M är antalet ASCII-tecken.
