Easy
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:
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.class Solution {
fun minTimeToType(word: String): Int {
var min = 0
var curr = 'a'
for (i in 0 until word.length) {
val diff = curr.code - word[i].code
curr = word[i]
min += Math.min(diff + 26, Math.min(Math.abs(diff), 26 - diff))
min++
}
return min
}
}