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: Clockwise or Counter-clockwise?
For each pair of (prev, curr), choose the shortest distance.
One is abs(p – c), another is 26 – abs(p – c).
Don’t forget to add 1 for typing itself.
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int minTimeToType(string word) { int ans = 0; char p = 'a'; for (char c : word) { ans += 1 + min(abs(c - p), 26 - abs(c - p)); p = c; } return ans; } }; |