Sunday, April 23, 2017

Lexicographically first palindromic string - GeeksforGeeks


Lexicographically first palindromic string - GeeksforGeeks
Rearrange the characters of the given string to form a lexicographically first palindromic string. If no such string exists display message "no palindromic string".

Properties for palindromic string:
1. If length of string is even, then the frequency of each character in the string must be even.
2. If the length is odd then there should be one character whose frequency is odd and all other chars must have even frequency and at-least one occurrence of the odd character must be present in the middle of the string.
Algorithm
1. Store frequency of each character in the given string
2. Check whether a palindromic string can be formed or not using the properties of palindromic string mentioned above.
3. If palindromic string cannot be formed, return “No Palindromic String”.
4. Else we create three strings and then return front_str + odd_str + rear_str.
  • odd_str : It is empty if there is no character with odd frequency. Else it contains all occurrences of odd character.
  • front_str : Contains half occurrences of all even occurring characters of string in increasing order.
  • rear_str Contains half occurrences of all even occurring characters of string in reverse order of front_str.
void countFreq(string str, int freq[], int len)
{
    for (int i=0; i<len; i++)
        freq[str.at(i) - 'a']++;
}
 
// Cases to check whether a palindr0mic
// string can be formed or not
bool canMakePalindrome(int freq[], int len)
{
    // count_odd to count no of
    // chars with odd frequency
    int count_odd = 0;
    for (int i=0; i<MAX_CHAR; i++)
        if (freq[i]%2 != 0)
            count_odd++;
 
    // For even length string
    // no odd freq character
    if (len%2 == 0)
    {
        if (count_odd > 0)
            return false;
        else
            return true;
    }
 
    // For odd length string
    // one odd freq character
    if (count_odd != 1)
        return false;
 
    return true;
}
 
// Function to find odd freq char and
// reducing its freq by 1returns "" if odd freq
// char is not present
string findOddAndRemoveItsFreq(int freq[])
{
    string odd_str = "";
    for (int i=0; i<MAX_CHAR; i++)
    {
        if (freq[i]%2 != 0)
        {
            freq[i]--;
            odd_str = odd_str + (char)(i+'a');
            return odd_str;
        }
    }
    return odd_str;
}
 
// To find lexicographically first palindromic
// string.
string findPalindromicString(string str)
{
    int len = str.length();
 
    int freq[MAX_CHAR] = {0};
    countFreq(str, freq, len);
 
    if (!canMakePalindrome(freq, len))
        return "No Palindromic String";
 
    // Assigning odd freq character if present
    // else empty string.
    string odd_str = findOddAndRemoveItsFreq(freq);
 
    string front_str = "", rear_str = " ";
 
    // Traverse characters in increasing order
    for (int i=0; i<MAX_CHAR; i++)
    {
        string temp = "";
        if (freq[i] != 0)
        {
            char ch = (char)(i + 'a');
 
            // Divide all occurrences into two
            // halves. Note that odd character
            // is removed by findOddAndRemoveItsFreq()
            for (int j=1; j<=freq[i]/2; j++)
                temp = temp + ch;
 
            // creating front string
            front_str = front_str + temp;
 
            // creating rear string
            rear_str = temp + rear_str;
        }
    }
 
    // Final palindromic string which is
    // lexicographically first
    return (front_str + odd_str + rear_str);
}
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