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.
brute force, O(rc), TLE:
class Solution(object):
def wordsTyping(self, sentence, rows, cols):
"""
:type sentence: List[str]
:type rows: int
:type cols: int
:rtype: int
"""
i=0
cnt=0
for r in range(rows):
col=cols
while col>=len(sentence[i]):
col-=len(sentence[i])+1
i+=1
if i==len(sentence):
cnt+=1
i=0
return cnt
O(r+len(sentence)), use dp to record next movement:
class Solution(object):
def wordsTyping(self, sentence, rows, cols):
"""
:type sentence: List[str]
:type rows: int
:type cols: int
:rtype: int
"""
s=' '.join(sentence)+' '
dp=[0]*len(s)
for i in range(1, len(dp)):
dp[i]=dp[i-1]-1 if s[i]!=' ' else 1
cnt=0
for r in range(rows):
cnt+=cols
cnt+=dp[cnt%len(s)]
return cnt/len(s)