Python C++速解Leetcode 2073 Time Needed to Buy Tickets

Python CPP速解LLeetcode 2073  Time Needed to Buy Tickets

假設x=tickets[k]。 當i≤k時，person[i]最多只能買x張票； 當 i大於k時，person[i]最多只能購買 x-1 票。python code請進

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Assume x=tickets[k]. When i≤k, person[i] can only buy x tickets at most; when i bigger than k, person[i] can only buy x-1 tickets at most.

[codes on Leetcode]https://leetcode.com/problems/time-needed-to-buy-tickets/solutions/4995373/loop-with-sum-min-buy-tickets-i-1-liner-0ms-beats-100/

[codes on Leetcode 678]https://leetcode.com/problems/valid-parenthesis-string/solutions/4985180/recursive-dp-iterative-dp-vs-greedy-0ms-beats-100/

[Prefix Sum playlist]https://www.youtube.com/watch?v=FGDkETbwWHc&list=PLYRlUBnWnd5LkYN-5ptHDrtPXOuHm9uSq&index=1

class Solution:
    def timeRequiredToBuy(self, t: List[int], k: int) -> int:
        x=t[k]
        time=sum(min(y,x) for y in t[:k+1])
        time+=sum(min(y, x-1) for y in t[k+1:])
        return time
        

Python 1-liner

class Solution:
    def timeRequiredToBuy(self, t: List[int], k: int) -> int:
        return sum(min(y,x:=t[k]) for y in t[:k+1])+sum(min(y, x-1) for y in t[k+1:])