Given a string, find the longest palindrome that can be constructed by removing or shuffling characters from the string. Return only one palindrome if there are multiple palindrome strings of longest length.
Examples:
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 can divide any palindromic string into three parts – beg, mid and end. For palindromic string of odd length say 2n + 1, ‘beg’ consists of first n characters of the string, ‘mid’ will consist of only 1 character i.e. (n + 1)th character and ‘end’ will consists of last n characters of the palindromic string. For palindromic string of even length 2n, ‘mid’ will always be empty. It should be noted that ‘end’ will be reverse of ‘beg’ in order for string to be palindrome.
The idea is to use above observation in our solution. As shuffling of characters is allowed, order of characters doesn’t matter in the input string. We first get frequency of each character in the input string. Then all characters having even occurrence (say 2n) in the input string will be part of the output string as we can easily place n characters in ‘beg’ string and the other n characters in the ‘end’ string (by preserving the palindromic order). For characters having odd occurrence (say 2n + 1), we fill ‘mid’ with one of all such characters. and remaining 2n characters are divided in halves and added at beginning and end.
Below is the implementation of above idea –
C++
// C++ program to find the longest palindrome by removing // or shuffling characters from the given string #include <bits/stdc++.h> 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 strring # Function to find the longest palindrome by removing # or shuffling characters from the given strring def findLongestPalindrome(strr): # to stores freq of characters in a strring count = [0]*256 # find freq of characters in the input strring 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 strring. 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 # strring 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 strring 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 |
Output:
abcdcba
Time complexity of above solution is O(n) where n is length of the string. Since, number of characters in the alphabet is constant, they do not contribute to asymptotic analysis.
Auxiliary space used by the program is M where M is number of ASCII characters.
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