CareerCup

1. You can make use of a Hashmap for each string. Each line is a string. Use a hashmap to count all the characters. If the length is odd, 1 character can appear once, rest all have to appear or the count has to be even for other characters. The characters that fall short of this condition are the number of characters required to make the string a palindrome. The you can just go over the hashmap, use two pointers front and rear to place the elements. Time complexity O(n) for the lines, O(T) for each string, So O(Tn) for the file containing all the strings.
2. Sort the string and count the characters. Follow similar approach as in case 1.
Time complexity O(n) for lines, O(TlogT) for sorting. O(nTlogT) in total.

// ZoomBA
def count_more_char( string  ){
  m = mset ( string.value )
  // find all chars with odd frequency 
  odds = select ( m ) :: { $.o.value % 2 != 0 }
  len = size(odds)
  // every char freq is even means 
  //can be shuffled into a palindrome
  if ( 0 == len ) return 0  
  // because only one odd freq can be allowed 
  //for palindrome, rest needs to be made into even 
  len - 1  
}

Based on expansion of WilOzil's idea.

first you can maked any app