Sentence screen fitting

class Solution:
    def wordsTyping(self, sentence: List[str], rows: int, cols: int) -> int:
        sen = " ".join(sentence) + " "
        n = len(sen)
        start = 0
        for row in range(rows):
            #new start point on next row after cols chars displayed 
            start = start + cols 
            if sen[start % n] == " ":
                start += 1
            else:
                # if after displaying cols chars, we reach middle of a word 
                # then we need to move to the start of the word and try to display
                # it on next line 
                while(start > 0 and sen[(start-1) % n] != " "):
                    start -= 1

        return start // n

Sentence Screen Fitting

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