There are n
people in a line queuing to buy tickets, where the 0th
person is at the front of the line and the (n - 1)th
person is at the back of the line.
You are given a 0-indexed integer array tickets
of length n
where the number of tickets that the ith
person would like to buy is tickets[i]
.
Each person takes exactly 1 second to buy a ticket. A person can only buy 1 ticket at a time and has to go back to the end of the line (which happens instantaneously) in order to buy more tickets. If a person does not have any tickets left to buy, the person will leave the line.
Return the time taken for the person at position k
(0-indexed) to finish buying tickets.
Example 1:
Input: tickets = [2,3,2], k = 2 Output: 6 Explanation: - In the first pass, everyone in the line buys a ticket and the line becomes [1, 2, 1]. - In the second pass, everyone in the line buys a ticket and the line becomes [0, 1, 0]. The person at position 2 has successfully bought 2 tickets and it took 3 + 3 = 6 seconds.
Example 2:
Input: tickets = [5,1,1,1], k = 0 Output: 8 Explanation: - In the first pass, everyone in the line buys a ticket and the line becomes [4, 0, 0, 0]. - In the next 4 passes, only the person in position 0 is buying tickets. The person at position 0 has successfully bought 5 tickets and it took 4 + 1 + 1 + 1 + 1 = 8 seconds.
Constraints:
n == tickets.length
1 <= n <= 100
1 <= tickets[i] <= 100
0 <= k < n
We can solve this by simple math. Add up the number of tickets needed for each person, the only thing needs to be aware is index if it’s before k then, adding it up if tickets[i] <= tickets[k] else tickets[k]. If it’s after k, then remember to minus 1.
class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
required_tickets = tickets[k]
time = 0
for i, ticket in enumerate(tickets):
if i > k:
time += ticket if ticket <= (required_tickets - 1) else (required_tickets - 1)
else:
time += ticket if ticket <= required_tickets else required_tickets
return time
搶先發佈留言