First non-repeating character in a stream.cpp

/*
First non-repeating character in a stream
==========================================

Given an input stream of A of n characters consisting only of lower case alphabets. The task is to find the first non repeating character, each time a character is inserted to the stream. If there is no such character then append '#' to the answer.

Example 1:
Input: A = "aabc"
Output: "a#bb"
Explanation: For every character first non
repeating character is as follow-
"a" - first non-repeating character is 'a'
"aa" - no non-repeating character so '#'
"aab" - first non-repeating character is 'b'
"aabc" - first non-repeating character is 'b'

Example 2:
Input: A = "zz"
Output: "z#"
Explanation: For every character first non
repeating character is as follow-
"z" - first non-repeating character is 'z'
"zz" - no non-repeating character so '#'

Your Task:
You don't need to read or print anything. Your task is to complete the function FirstNonRepeating() which takes A as input parameter and returns a string after processing the input stream.

Expected Time Complexity: O(26 * n)
Expected Space Complexity: O(26)

Constraints:
1 <= n <= 105
*/

string FirstNonRepeating(string A)
{
  list<char> dll, temp;
  vector<int> repeated(26, 0);
  vector<list<char>::iterator> ptrs(26, temp.begin());
  string ans;

  for (auto &i : A)
  {
    if (!repeated[i - 'a'] && ptrs[i - 'a'] == temp.begin())
    {
      dll.push_back(i);
      auto it = dll.end();
      it--;
      ptrs[i - 'a'] = it;
    }
    else if (!repeated[i - 'a'] && ptrs[i - 'a'] != temp.begin())
    {
      dll.erase(ptrs[i - 'a']);
      repeated[i - 'a'] = 1;
    }

    if (dll.size())
      ans.push_back(dll.front());
    else
      ans.push_back('#');
  }
  return ans;
}