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Tested Python solution for LeetCode 418 with 16 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 418 # by problem number
lcpy gen -s sentence_screen_fitting # by problem name
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.
Input: sentence = ["hello","world"], rows = 2, cols = 8
Output: 1
Explanation:
hello---
world---
The character '-' signifies an empty space on the screen.
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.
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.
1 <= sentence.length1 <= rows, cols <= 4 * 10^41 <= sentence[i].length <= 10sentence[i] consists of only lower-case English letters.class Solution:
# Time: O(rows + total_sentence_length)
# Space: O(total_sentence_length) for the joined sentence string
def words_typing(self, sentence: list[str], rows: int, cols: int) -> int:
s = " ".join(sentence) + " "
n = len(s)
total = 0
for _ in range(rows):
total += cols
if s[total % n] == " ":
total += 1
else:
while total > 0 and s[(total - 1) % n] != " ":
total -= 1
return total // n
| Time | Space |
|---|---|
| O(rows + total_sentence_length) | O(total_sentence_length) for the joined sentence string |