Find longest palindrome formed by removing or shuffling chars from string - GeeksforGeeks

Find longest palindrome formed by removing or shuffling chars from string - GeeksforGeeks
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.

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.

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.

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;

}

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