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• 
  

  1

  Avocadorable    2:05 p.m. 7 Tháng 6, 2024

  from collections import defaultdict
  
  
  def getPalindrome(st):
  
      # Store counts of characters
      hmap = defaultdict(int)
      for i in range(len(st)):
          hmap[st[i]] += 1
  
      # Find the number of odd elements.
      # Takes O(n)
      oddCount = 0
  
      for x in hmap:
          if (hmap[x] % 2 != 0):
              oddCount += 1
              oddChar = x
  
      # odd_cnt = 1 only if the length of
      # str is odd
      if (oddCount > 1 or oddCount == 1 and
              len(st) % 2 == 0):
          return "NO SOLUTION"
  
      # Generate first half of palindrome
      firstHalf = ""
      secondHalf = ""
  
      for x in sorted(hmap.keys()):
  
          # Build a string of floor(count/2)
          # occurrences of current character
          s = (hmap[x] // 2) * x
  
          # Attach the built string to end of
          # and begin of second half
          firstHalf = firstHalf + s
          secondHalf = s + secondHalf
  
      # Insert odd character if there
      # is any
      if (oddCount == 1):
          return (firstHalf + oddChar + secondHalf)
      else:
          return (firstHalf + secondHalf)
  
  
  # Driver code
  if __name__ == "__main__":
  
      s = input()
  
      print(getPalindrome(s))
  

• 
  

  0

  Vodangngoclam    5:45 p.m. 7 Tháng 6, 2024

  gà:
  

  Delphi

  var
      a : ansistring;
      c : array[65..90] of int64;
      le , chan : int64;
      i , j : longint;
  
  BEGIN
      read(a);
      For i:=1 to length(a) do inc(c[ord(a[i])]);
      For i:=65 to 90 do
          IF (c[i] <> 0) THEN
              IF c[i] mod 2 = 0 THEN
                  inc(chan)
              ELSE
                  inc(le);
      IF le > 1 THEN
          write('NO SOLUTION')
      ELSE
          BEGIN
              For i:=65 to 90 do
                  For j := 1 to c[i] div 2 do write(chr(i));
              IF le = 1 THEN
                  For i:=65 to 90 do
                      IF c[i] mod 2 = 1 THEN
                          write(chr(i));
              For i:=90 downto 65 do
                  For j:=1 to c[i] div 2 do write(chr(i));
          END;
      readln;
  END.
  

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  0

  PhucDepZai    10:50 a.m. 10 Tháng 7, 2024

  cũng đúng mà , có sai đâu
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