Sunday, October 9, 2016

LeetCode 418 - Sentence Screen Fitting


http://bookshadow.com/weblog/2016/10/09/leetcode-sentence-screen-fitting/
Given a rows x cols screen and a sentence represented by a list of 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 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.

https://discuss.leetcode.com/topic/62455/21ms-18-lines-java-solution
  1. String s = String.join(" ", sentence) + " " ;. This line gives us a formatted sentence to be put to our screen.
  2. start is the counter for how many valid characters from s have been put to our screen.
  3. if (s.charAt(start % l) == ' ') is the situation that we don't need an extra space for current row. The current row could be successfully fitted. So that we need to increase our counter by using start++.
  4. The else is the situation, which the next word can't fit to current row. So that we need to remove extra characters from next word.
  5. start / s.length() is (# of valid characters) / our formatted sentence.
    public int wordsTyping(String[] sentence, int rows, int cols) {
        String s = String.join(" ", sentence) + " ";
        int start = 0, l = s.length();
        for (int i = 0; i < rows; i++) {
            start += cols;
            if (s.charAt(start % l) == ' ') {
                start++;
            } else {
                while (start > 0 && s.charAt((start-1) % l) != ' ') {
                    start--;
                }
            }
        }
        
        return start / s.length();
    }

https://discuss.leetcode.com/topic/62364/java-optimized-solution-17ms
public int wordsTyping(String[] sentence, int rows, int cols) {
        int[] nextIndex = new int[sentence.length];
        int[] times = new int[sentence.length];
        for(int i=0;i<sentence.length;i++) {
            int curLen = 0;
            int cur = i;
            int time = 0;
            while(curLen + sentence[cur].length() <= cols) {
                curLen += sentence[cur++].length()+1;
                if(cur==sentence.length) {
                    cur = 0;
                    time ++;
                }
            }
            nextIndex[i] = cur;
            times[i] = time;
        }
        int ans = 0;
        int cur = 0;
        for(int i=0; i<rows; i++) {
            ans += times[cur];
            cur = nextIndex[cur];
        }
        return ans;
    }

https://discuss.leetcode.com/topic/62297/a-simple-simulation
http://bookshadow.com/weblog/2016/10/09/leetcode-sentence-screen-fitting/
上例中apple单词的相对位置从第二行开始循环,因此只需要找到单词相对位置的“循环节”,即可将问题简化。
利用字典dp记录循环节的起始位置,具体记录方式为:dp[(pc, pw)] = pr, ans
以数对(pc, pw)为键,其中pw为单词在句子中出现时的下标,pc为单词出现在屏幕上的列数

以数对(pr, ans)为值,其中pr为单词出现在屏幕上的行数,ans为此时已经出现过的完整句子数

def wordsTyping(self, sentence, rows, cols): """ :type sentence: List[str] :type rows: int :type cols: int :rtype: int """ wcount = len(sentence) wlens = map(len, sentence) slen = sum(wlens) + wcount dp = dict() pr = pc = pw = ans = 0 while pr < rows: if (pc, pw) in dp: pr0, ans0 = dp[(pc, pw)] loop = (rows - pr0) / (pr - pr0 + 1) ans = ans0 + loop * (ans - ans0) pr = pr0 + loop * (pr - pr0) else: dp[(pc, pw)] = pr, ans scount = (cols - pc) / slen ans += scount pc += scount * slen + wlens[pw] if pc <= cols: pw += 1 pc += 1 if pw == wcount: pw = 0 ans += 1 if pc >= cols: pc = 0 pr += 1 return ans

Brute Force
http://www.itdadao.com/articles/c15a559599p0.html
    public int wordsTyping(String[] sentence, int rows, int cols) 
    {
        int m = sentence.length;
        int res = 0;
        int c = 0;
        int left = cols;
        for(int i = 0; i < rows;)
        {
            if(sentence[c%m].length() <= left)
            {
                if(c%m == m-1) res++;
                c++;
                left = left - sentence[(c-1)%m].length() - 1;
                
            }
            else
            {
                i++;
                left = cols;
            }
        }
        return res;        
    }



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