Leetcode:sentence Screen Fitting

Given a rows x cols screens and a sentence represented by a list of words, find how many times the Given sentence can B E fitted on the screen. Note:a Word cannot is split into the lines. The order of words in the sentence must remain unchanged. The consecutive words in a line must is separated by a single space. Total words in the sentence won' t exceed 100.Length of each word won ' t exceed 10.1≤rows, cols≤20,000. Example1: Input:rows= 2, cols = 8, sentence = ["Hello", "World"]output:1Explanation:hello--- World---The character‘-‘signifies an empty space in the screen. Example2: Input:rows= 3, cols = 6, sentence = ["A", "BCD", "E"]output:2explanation:a-bcd-e-A---BCD-e-The character‘-‘signifies an empty space in the screen. Example3: Input:rows= 4, cols = 5, sentence = ["I", "had", "apple", "pie"]output:1explanation:i-Hadapplepie-Ihad--The character'-' signifies an empty space in the screen.

First a brute force, similar to the text adjustment

1  Public classSolution {2      Public intWordstyping (string[] sentence,intRowsintcols) {3         if(sentence==NULL|| sentence.length==0 | | Sentence.length>rows*cols | | rows<=0 | | Cols<=0)4             return0;5         intres = 0;6         intj = 0;//indicate the index of string in sentence that's currently trying to being inserted to current row7         introw = 0;//Current Row8         intcol = 0;//Current Col9         Ten          while(Row <rows) { One              while(col + sentence[j].length ()-1 <cols) { ACol = col + sentence[j].length () + 1; -J + +; -                 if(J = =sentence.length) { theres++; -j = 0; -                 } -             } +row++; -Col = 0; +         } A         returnRes; at     } -}

But in a slightly larger case, the tle, for example:

["A", "B", "E"] 20000 20000, spent 465ms

So think about how to save time,

Hint is can DP, think how to reuse, refer to:https://discuss.leetcode.com/topic/62364/java-optimized-solution-17ms

If this line starts with a string in sentence, then which string begins the next line, which is OK, and if the bank will not reach the end of sentence, if so, to a few times, this is a definite

These two points can be exploited because we have found the DP's multiplexing relationship

Sub-problem:if there ' s a new line which is starting with certain index in sentence, what is the starting index of next Li NE (nextindex[]). BTW, we compute how many times the "pointer in" on the current line passes over the last index (times[]).

Time Complexity:o (n (cols/lenaverage)) + O (rows), where n is the length of sentence array, lenaverage is the average Len Gth of the words in the input array.

1  Public classSolution {2      Public intWordstyping (string[] sentence,intRowsintcols) {3         int[] Nextint =New int[sentence.length];4         int[] times =New int[sentence.length];5          for(inti=0; i<sentence.length; i++) {6             intcur = i;//try to insert string with index cur in sentence to current row7             intcol = 0;//Current Col8             intTime = 0;9              while(col + sentence[cur].length ()-1 <cols) {TenCol = col + sentence[cur++].length () + 1; One                 if(cur = =sentence.length) { ACur = 0; -time++; -                 } the             } -Nextint[i] =cur; -Times[i] =Time ; -         } +          -         intres = 0; +         intCur = 0; A          for(inti=0; i<rows; i++) { atRes + =Times[cur]; -Cur =Nextint[cur]; -         } -         returnRes; -     } -}

Leetcode:sentence Screen Fitting