Atsižvelgdami į eilutę, suraskite ilgiausią palindromą, kurį galima sukurti pašalinant arba maišant simbolius iš eilutės. Grąžinkite tik vieną palindromą, jei yra kelios ilgiausio ilgio palindromo eilutės.
Pavyzdžiai:
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
Bet kurią palindrominę eilutę galime padalyti į tris dalis – beg mid ir end. Nelyginio ilgio palindrominės eilutės atveju tarkime, kad 2n + 1 'prašymas' susideda iš pirmųjų n eilutės simbolių, 'vidurį' sudarys tik 1 simbolis, t. Vienodo ilgio palindrominei stygai 2n 'viduris' visada bus tuščias. Reikėtų pažymėti, kad „pabaiga“ bus priešinga „pradėti“, kad eilutė būtų palindrominė.
Idėja yra naudoti aukščiau pateiktą stebėjimą mūsų sprendime. Kadangi leidžiama maišyti simbolius, simbolių tvarka įvesties eilutėje neturi reikšmės. Pirmiausia gauname kiekvieno simbolio dažnį įvesties eilutėje. Tada visi simboliai, kurių atsiradimas yra lygus (tarkime, 2n) įvesties eilutėje, bus išvesties eilutės dalis, nes mes galime lengvai įdėti n simbolių į eilutę „pradeda“, o kitus n simbolius „pabaigos“ eilutėje (išsaugodami palindrominę tvarką). Nelyginių simbolių atveju (tarkim 2n + 1) užpildome „vidurį“ vienu iš visų tokių simbolių. o likę 2n simboliai padalijami į pusę ir pridedami pradžioje ir pabaigoje.
Žemiau yra aukščiau pateiktos idėjos įgyvendinimas
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>
Išvestis
abcdcba
Laiko sudėtingumas aukščiau pateiktas sprendimas yra O(n), kur n yra eilutės ilgis. Kadangi abėcėlės simbolių skaičius yra pastovus, jie neprisideda prie asimptotinės analizės.
Pagalbinė erdvė programos naudojamas M, kur M yra ASCII simbolių skaičius.