Lexicographically first palindromic string

2.2

Rearrange the characters of the given string to form a lexicographically first palindromic string. If no such string exists display message “no palindromic string”.

Examples:

Input : malayalam
Output : aalmymlaa

Input : apple
Output : no palindromic string

Simple Approach:
1. Sort the string characters in alphabetical(ascending) order.
2. One be one find lexicographically next permutation of the given string.
3. The first permutation which is palindrome is the answer.

Efficient Approach: 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.
    • Below is implementation of above steps.

      // C++ program to find first palindromic permutation
      // of given string
      #include <bits/stdc++.h>
      using namespace std;
      
      const char MAX_CHAR = 26;
      
      // Function to count frequency of each char in the
      // string. freq[0] for 'a',...., freq[25] for 'z'
      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);
      }
      
      // Driver program
      int main()
      {
          string str = "malayalam";
          cout << findPalindromicString(str);
          return 0;
      }
      

      Output:

      aalmymlaa
      

      Time Complexity : O(n) where n is length of input string. Assuming that size of string alphabet is constant.

      This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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