418. 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.

S： 记忆化搜索 O(col*row)

map[i]存储由sentence[i]开始的一行最多可以排多少个词

class Solution {
public:
    int wordsTyping(vector<string>& sentence, int rows, int cols) {
        int size = sentence.size();
        int count = 0;
        vector<int> map(size, 0);
        for (int i = 0; i < rows; ++i) {
            int start = count % size;
            if (map[start]) {
                count += map[start];
            } else {
                int newCount = 0;
                for (int j = 0, idx = start; j < cols; 
                     j += sentence[idx].size() + 1, idx = (idx + 1) % size) {
                    if (j + sentence[idx].size() > cols) break;
                    newCount++;
                }
                map[start] = newCount;
                count += newCount;
            }
        }
        return count / size;
    }
};