Maximum number of characters using keystrokes A, Ctrl+A, Ctrl+C and Ctrl+V

Question

This is an interview question from google. I am not able to solve it by myself. Can somebody shed some light?

Write a program to print the sequence of keystrokes suc

Answer 1

By using marcog's solution I found a pattern that starts at n=16. To illustrate this here are the keystrokes for n=24 up to n=29, I replaced ^A with S (select), ^C with C (copy), and ^V with P (paste) for readability:

24: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P
       4   *    4    *    4    *    4    *    4     = 1024
25: A,A,A,A,S,C,P,P,P,S,C,P,P,S,C,P,P,S,C,P,P,S,C,P,P
       4   *    4    *   3   *   3   *   3   *   3    = 1296
26: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,S,C,P,P,S,C,P,P
       4   *    4    *    4    *   3   *   3   *   3    = 1728
27: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,S,C,P,P
       4   *    4    *    4    *    4    *   3   *   3    = 2304
28: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P
       4   *    4    *    4    *    4    *    4    *   3    = 3072
29: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P
       4   *    4    *    4    *    4    *    4    *    4     = 4096

After an initial 4 As, the ideal pattern is to select, copy, paste, paste, paste and repeat. This will multiply the number of As by 4 every 5 keystrokes. If this 5 keystroke pattern cannot consume the remaining keystrokes on its own some number of 4 keystroke patterns (SCPP) consume the final keystrokes, replacing SCPPP (or removing one of the pastes) as necessary. The 4 keystroke patterns multiply the total by 3 every 4 keystrokes.

Using this pattern here is some Python code that gets the same results as marcog's solution, but is O(1) edit: This is actually O(log n) due to exponentiation, thanks to IVlad for pointing that out.

def max_chars(n):
  if n <= 15:
    return (0, 1, 2, 3, 4, 5, 6, 9, 12, 16, 20, 27, 36, 48, 64, 81)[n]
  e3 = (4 - n) % 5
  e4 = n // 5 - e3
  return 4 * (4 ** e4) * (3 ** e3)

Calculating e3: There are always between 0 and 4 SCPP patterns at the end of the keystroke list, for n % 5 == 4 there are 4, n % 5 == 1 there are 3, n % 5 == 2 there are 2, n % 5 == 3 there are 1, and n % 5 == 4 there are 0. This can be simplified to (4 - n) % 5.

Calculating e4: The total number of patterns increases by 1 whenever n % 5 == 0, as it turns out this number increases to exactly n / 5. Using floor division we can get the total number of patterns, the total number for e4 is the total number of patterns minus e3. For those unfamiliar with Python, // is the future-proof notation for floor division.