// 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 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>