# Sentence Screen Fitting

Given a`rows x cols`screen and a sentence represented by a list of**non-empty**words, find**how many times**the given sentence can be fitted on the screen.

**Note:**

1. A word cannot be split into two lines.
2. The order of words in the sentence must remain unchanged.
3. Two consecutive words

   **in a line**

   must be separated by a single space.
4. Total words in the sentence won't exceed 100.
5. Length of each word is greater than 0 and won't exceed 10.
6. 1 ≤ rows, cols ≤ 20,000.

**Example 1:**

```
Input:

rows = 2, cols = 8, sentence = ["hello", "world"]


Output:

1


Explanation:

hello---
world---

The character '-' signifies an empty space on the screen.
```

**Example 2:**

```
Input:

rows = 3, cols = 6, sentence = ["a", "bcd", "e"]


Output:

2


Explanation:

a-bcd- 
e-a---
bcd-e-

The character '-' signifies an empty space on the screen.
```

**Example 3:**

```
Input:

rows = 4, cols = 5, sentence = ["I", "had", "apple", "pie"]


Output:

1


Explanation:

I-had
apple
pie-I
had--

The character '-' signifies an empty space on the screen.
```

分析

start表示当前总放入字符数。 每次row直接start+= cols。 然后start%len是空格代表本行不需要Padding。 不是的话去掉最后单词。

```
class Solution:
    def wordsTyping(self, sentence: List[str], rows: int, cols: int) -> int:
        s = ' '.join(sentence)+' '
        ln = len(s)
        start = 0
        for i in range(rows):
            start += cols
            if s[start %ln] == ' ': #可以塞入当前row,不用padding
                start += 1
            else:#去掉最后一个word,去下一行
                while start > 0 and s[(start-1)%ln] != ' ':#loop出来start在某word第一个字母处
                    start -= 1
        return  start//ln
```


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