# 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:
apple
pie-i
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

• 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);
}