Minimum Time to Type Word Using Special Typewriter

There is a special typewriter with lowercase English letters 'a' to 'z' arranged in a circle with a pointer. A character can only be typed if the pointer is pointing to that character. The pointer is initially pointing to the character 'a'. Each second, you may perform one of the following operations:

• Move the pointer one character counterclockwise or clockwise.
• Type the character the pointer is currently on.

Given a string word, return the minimum number of seconds to type out the characters in word.

Example 1:

Input: word = "abc" Output: 5

Example 2:

Input: word = "bza" Output: 7

Example 3:

Input: word = "zjpc" Output: 34

Constraints:

• 1 <= word.length <= 100
• word consists of lowercase English letters.

def min_time_to_type(word: str) -> int:
    """Calculates the minimum time to type a word on a circular typewriter.

    Args:
        word: The word to type.

    Returns:
        The minimum number of seconds to type the word.
    """
    current = 'a'
    total_time = 0
    for char in word:
        diff = abs(ord(char) - ord(current))
        move_time = min(diff, 26 - diff)
        total_time += move_time + 1  # 1 second to type the character
        current = char
    return total_time

# Example Usage
word1 = "abc"
print(f"Input: {word1}, Output: {min_time_to_type(word1)}")

word2 = "bza"
print(f"Input: {word2}, Output: {min_time_to_type(word2)}")

word3 = "zjpc"
print(f"Input: {word3}, Output: {min_time_to_type(word3)}")

Explanation:

1. Initialization:

   • current = 'a': The pointer starts at 'a'.
   • total_time = 0: Initialize the total time taken.

2. Iteration:

   • The code iterates through each character char in the input word.
   • diff = abs(ord(char) - ord(current)): Calculates the absolute difference between the ASCII values of the current character and the pointer's current position.
   • move_time = min(diff, 26 - diff): Determines the minimum time to move the pointer to the required character. Since the characters are arranged in a circle, we take the minimum of the clockwise and counterclockwise distances.
   • total_time += move_time + 1: Adds the move_time and 1 (for typing the character) to the total_time.
   • current = char: Updates the pointer's current position to the character that was just typed.

3. Return Value:

   • The function returns the total_time, which represents the minimum time required to type the entire word.

Big O Analysis:

• Time Complexity: O(n)

  • The code iterates through each character in the input word once. The operations inside the loop (calculating the difference, finding the minimum, and adding to the total time) take constant time, O(1).
  • Therefore, the overall time complexity is directly proportional to the length of the word, which is O(n).

• Space Complexity: O(1)

  • The code uses a constant amount of extra space, regardless of the input size. The variables current, total_time, diff, and move_time take up constant space.
  • Therefore, the space complexity is O(1).

Edge Cases:

1. Empty Word:

   • If the input word is empty, the loop will not execute, and the function will return 0, which is the correct behavior.

2. Word with Same Characters:

   • If the input word contains repeated characters, the move_time might be 0 for some characters. The code handles this case correctly.

3. Word with All 'a's:

   • If the input word consists only of 'a's, the pointer will never need to move, and the time will be equal to the length of the word.

4. Long Word:

   • The code correctly handles long words by iterating through each character and calculating the minimum move time.

5. Characters outside 'a' to 'z':

   • The code assumes that the input only contains lower-case english letters, and will produce unexpected behavior if it is violated.