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.
Similar Questions:
// OJ: https://leetcode.com/problems/minimum-time-to-type-word-using-special-typewriter/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
int minTimeToType(string s) {
int ans = 0, prev = 'a';
for (char c : s) {
int diff = abs(c - prev);
ans += min(diff, 26 - diff);
prev = c;
++ans;
}
return ans;
}
};