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418. Sentence Screen Fitting
Description
Given a rows x cols screen and a sentence represented as a list of strings, return the number of times the given sentence can be fitted on the screen.
The order of words in the sentence must remain unchanged, and a word cannot be split into two lines. A single space must separate two consecutive words in a line.
Example 1:
Input: sentence = ["hello","world"], rows = 2, cols = 8 Output: 1 Explanation: hello--- world--- The character '-' signifies an empty space on the screen.
Example 2:
Input: sentence = ["a", "bcd", "e"], rows = 3, cols = 6 Output: 2 Explanation: a-bcd- e-a--- bcd-e- The character '-' signifies an empty space on the screen.
Example 3:
Input: sentence = ["i","had","apple","pie"], rows = 4, cols = 5 Output: 1 Explanation: i-had apple pie-i had-- The character '-' signifies an empty space on the screen.
Constraints:
1 <= sentence.length <= 1001 <= sentence[i].length <= 10sentence[i]consists of lowercase English letters.1 <= rows, cols <= 2 * 104
Solutions
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class Solution { public int wordsTyping(String[] sentence, int rows, int cols) { String s = String.join(" ", sentence) + " "; int m = s.length(); int cur = 0; while (rows-- > 0) { cur += cols; if (s.charAt(cur % m) == ' ') { ++cur; } else { while (cur > 0 && s.charAt((cur - 1) % m) != ' ') { --cur; } } } return cur / m; } } -
class Solution { public: int wordsTyping(vector<string>& sentence, int rows, int cols) { string s; for (auto& t : sentence) { s += t; s += " "; } int m = s.size(); int cur = 0; while (rows--) { cur += cols; if (s[cur % m] == ' ') { ++cur; } else { while (cur && s[(cur - 1) % m] != ' ') { --cur; } } } return cur / m; } }; -
class Solution: def wordsTyping(self, sentence: List[str], rows: int, cols: int) -> int: s = " ".join(sentence) + " " m = len(s) cur = 0 for _ in range(rows): cur += cols if s[cur % m] == " ": cur += 1 while cur and s[(cur - 1) % m] != " ": cur -= 1 return cur // m -
func wordsTyping(sentence []string, rows int, cols int) int { s := strings.Join(sentence, " ") + " " m := len(s) cur := 0 for i := 0; i < rows; i++ { cur += cols if s[cur%m] == ' ' { cur++ } else { for cur > 0 && s[(cur-1)%m] != ' ' { cur-- } } } return cur / m } -
function wordsTyping(sentence: string[], rows: number, cols: number): number { const s = sentence.join(' ') + ' '; let cur = 0; const m = s.length; for (let i = 0; i < rows; ++i) { cur += cols; if (s[cur % m] === ' ') { ++cur; } else { while (cur > 0 && s[(cur - 1) % m] !== ' ') { --cur; } } } return Math.floor(cur / m); }