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_418.java
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_418.java
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package com.fishercoder.solutions;
/**
* 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:
A word cannot be split into two lines.
The order of words in the sentence must remain unchanged.
Two consecutive words in a line must be separated by a single space.
Total words in the sentence won't exceed 100.
Length of each word is greater than 0 and won't exceed 10.
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.
*/
public class _418 {
public static class Solution1 {
/**
* credit: https://discuss.leetcode.com/topic/62455/21ms-18-lines-java-solution
* <p>
* 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;
int 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();
}
}
}