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1974. Minimum Time to Type Word Using Special Typewriter

Description

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.

Solutions

Solution 1: Greedy

We initialize the answer variable \(\textit{ans}\) to the length of the string, indicating that we need at least \(\textit{ans}\) seconds to type the string.

Next, we traverse the string. For each character, we calculate the minimum distance between the current character and the previous character, and add this distance to the answer. Then we update the current character to the previous character and continue traversing.

The time complexity is \(O(n)\), where \(n\) is the length of the string. The space complexity is \(O(1)\).

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class Solution:
    def minTimeToType(self, word: str) -> int:
        ans, a = len(word), ord("a")
        for c in map(ord, word):
            d = abs(c - a)
            ans += min(d, 26 - d)
            a = c
        return ans

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class Solution {
    public int minTimeToType(String word) {
        int ans = word.length();
        char a = 'a';
        for (char c : word.toCharArray()) {
            int d = Math.abs(a - c);
            ans += Math.min(d, 26 - d);
            a = c;
        }
        return ans;
    }
}

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class Solution {
public:
    int minTimeToType(string word) {
        int ans = word.length();
        char a = 'a';
        for (char c : word) {
            int d = abs(a - c);
            ans += min(d, 26 - d);
            a = c;
        }
        return ans;
    }
};

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func minTimeToType(word string) int {
    ans := len(word)
    a := rune('a')
    for _, c := range word {
        d := int(max(a-c, c-a))
        ans += min(d, 26-d)
        a = c
    }
    return ans
}

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function minTimeToType(word: string): number {
    let a = 'a'.charCodeAt(0);
    let ans = word.length;
    for (const c of word) {
        const d = Math.abs(c.charCodeAt(0) - a);
        ans += Math.min(d, 26 - d);
        a = c.charCodeAt(0);
    }
    return ans;
}

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