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418. Sentence Screen Fitting πŸ”’

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 <= 100
  • 1 <= sentence[i].length <= 10
  • sentence[i] consists of lowercase English letters.
  • 1 <= rows, cols <= 2 * 104

Solutions

Solution 1

Thinking

Testing word by word on each row is clumsy: there can be \(10^4\) rows and the sentence repeats.

Join the words with a trailing space into a cyclic string \(s\), and let \(\textit{cur}\) be the number of characters already covered. Each row adds \(\textit{cols}\): if that landing is a space, the row ended on a word boundary and we advance one more; otherwise we rewind to the previous space, i.e. push the unfinished word to the next row.

The number of full sentences is \(\lfloor \textit{cur}/|s| \rfloor\). The trailing space encodes the mandatory gap, and the rewind forbids splitting a word.

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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

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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;
    }
}

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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;
    }
};

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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
}

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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);
}

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