[Leet Code] Minimum Time to Type Word Using Special Typewriter
2 min readAug 21, 2021
Leet Code: https://leetcode.com/problems/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
Explanation:
The characters are printed as follows:
- Type the character 'a' in 1 second since the pointer is initially on 'a'.
- Move the pointer clockwise to 'b' in 1 second.
- Type the character 'b' in 1 second.
- Move the pointer clockwise to 'c' in 1 second.
- Type the character 'c' in 1 second.
Example 2:
Input: word = "bza"
Output: 7
Explanation:
The characters are printed as follows:
- Move the pointer clockwise to 'b' in 1 second.
- Type the character 'b' in 1 second.
- Move the pointer counterclockwise to 'z' in 2 seconds.
- Type the character 'z' in 1 second.
- Move the pointer clockwise to 'a' in 1 second.
- Type the character 'a' in 1 second.
Example 3:
Input: word = "zjpc"
Output: 34
Explanation:
The characters are printed as follows:
- Move the pointer counterclockwise to 'z' in 1 second.
- Type the character 'z' in 1 second.
- Move the pointer clockwise to 'j' in 10 seconds.
- Type the character 'j' in 1 second.
- Move the pointer clockwise to 'p' in 6 seconds.
- Type the character 'p' in 1 second.
- Move the pointer counterclockwise to 'c' in 13 seconds.
- Type the character 'c' in 1 second.
Constraints:
1 <= word.length <= 100
word
consists of lowercase English letters.
Solution:
class Solution(object):
def minTimeToType(self, word):
"""
:type word: str
:rtype: int
"""
counter = 0
previous_value = "a"
for i in word:
counter += min(self.clockwise(i, previous_value), 26 - self.clockwise(i,previous_value))
previous_value = i
counter += 1
return counter
def clockwise(self, letter, previous_value):
return abs(ord(letter) - ord(previous_value))