Gegeven een string, zoek het langste palindroom dat kan worden geconstrueerd door tekens uit de string te verwijderen of te schudden. Retourneert slechts één palindroom als er meerdere palindroomreeksen met de langste lengte zijn.
Voorbeelden:
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
We kunnen elke palindroomsnaar in drie delen verdelen: begin midden en eind. Voor een palindromische reeks met een oneven lengte, bijvoorbeeld 2n + 1, bestaat 'beg' uit de eerste n tekens van de reeks. 'mid' zal uit slechts 1 teken bestaan, d.w.z. (n + 1)e teken en 'end' zal bestaan uit de laatste n tekens van de palindromische reeks. Voor palindroomreeksen van gelijke lengte zal 2n 'mid' altijd leeg zijn. Opgemerkt moet worden dat 'end' het omgekeerde is van 'beg', zodat string een palindroom is.
Het idee is om bovenstaande observatie in onze oplossing te gebruiken. Omdat het schudden van karakters is toegestaan, doet de volgorde van de karakters er niet toe in de invoerreeks. We krijgen eerst de frequentie van elk teken in de invoerreeks. Dan zullen alle tekens die even voorkomen (bijvoorbeeld 2n) in de invoerreeks deel uitmaken van de uitvoerreeks, omdat we gemakkelijk n tekens in de 'beg'-reeks kunnen plaatsen en de andere n tekens in de 'end'-reeks (door de palindromische volgorde te behouden). Voor tekens die oneven voorkomen (bijvoorbeeld 2n + 1) vullen we 'midden' met een van al dergelijke tekens. en de resterende 2n tekens worden in twee helften verdeeld en aan het begin en einde toegevoegd.
Hieronder ziet u de implementatie van bovenstaand idee
C++
// C++ program to find the longest palindrome by removing // or shuffling characters from the given string #include
// 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 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# 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
<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>
Uitvoer
abcdcba
Tijdcomplexiteit van bovenstaande oplossing is O(n) waarbij n de lengte van de string is. Omdat het aantal tekens in het alfabet constant is, dragen ze niet bij aan asymptotische analyse.
Hulpruimte gebruikt door het programma is M waarbij M het aantal ASCII-tekens is.