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Formatted question description: https://leetcode.ca/all/1165.html
1165. Single-Row Keyboard
Level
Easy
Description
There is a special keyboard with all keys in a single row.
Given a string keyboard
of length 26 indicating the layout of the keyboard (indexed from 0 to 25), initially your finger is at index 0. To type a character, you have to move your finger to the index of the desired character. The time taken to move your finger from index i
to index j is |i - j|
.
You want to type a string word
. Write a function to calculate how much time it takes to type it with one finger.
Example 1:
Input: keyboard = “abcdefghijklmnopqrstuvwxyz”, word = “cba”
Output: 4
Explanation: The index moves from 0 to 2 to write ‘c’ then to 1 to write ‘b’ then to 0 again to write ‘a’.
Total time = 2 + 1 + 1 = 4.
Example 2:
Input: keyboard = “pqrstuvwxyzabcdefghijklmno”, word = “leetcode”
Output: 73
Constraints:
keyboard.length == 26
keyboard
contains each English lowercase letter exactly once in some order.1 <= word.length <= 10^4
word[i]
is an English lowercase letter.
Solution
Initially the index is 0. For each letter in word
, find out the index of the letter in keyboard
, calculate the time taken to move the finger, and add the time taken to the total time. Finally, return the total time.
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class Solution { public int calculateTime(String keyboard, String word) { int time = 0; int index = 0; int length = word.length(); for (int i = 0; i < length; i++) { char c = word.charAt(i); int curIndex = keyboard.indexOf(c); time += Math.abs(curIndex - index); index = curIndex; } return time; } }
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// OJ: https://leetcode.com/problems/single-row-keyboard/ // Time: O(N) // Space: O(1) class Solution { public: int calculateTime(string keyboard, string word) { unordered_map<char, int> m; for (int i = 0; i < 26; ++i) m[keyboard[i]] = i; int prev = 0, ans = 0; for (char c : word) { int i = m[c]; ans += abs(prev - i); prev = i; } return ans; } };
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class Solution: def calculateTime(self, keyboard: str, word: str) -> int: index = {c: i for i, c in enumerate(keyboard)} res = t = 0 for c in word: res += abs(index[c] - t) t = index[c] return res
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func calculateTime(keyboard string, word string) int { index := map[byte]int{} for i := 0; i < len(keyboard); i++ { index[keyboard[i]] = i } res := 0 t := 0 for i := 0; i < len(word); i++ { res += abs(index[word[i]] - t) t = index[word[i]] } return res } func abs(x int) int { if x < 0 { return -x } return x }