> For the complete documentation index, see [llms.txt](https://cai-sen-se.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://cai-sen-se.gitbook.io/leetcode/2001-3000/2073.-mai-piao-xu-yao-de-shi-jian.md).

# 2073.买票需要的时间

<https://leetcode-cn.com/problems/time-needed-to-buy-tickets/>

## 观察法

对k及k之前的每个人i，若i买的票比k少，则i要的票一定能被满足；否则i最多能买到和k一样多的票

对k之后的每个人i，若i买的票大于等于k，则最多能满足`tickets[k] - 1`（因为k被满足后就结束了）；否则i要的票能被满足

```python
class Solution:
    def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
        n = len(tickets)
        cnt = 0
        for i in range(n):
            if i <= k:
                if tickets[i] < tickets[k]:
                    cnt += tickets[i]
                else:
                    cnt += tickets[k]
            elif i > k:
                if tickets[i] >= tickets[k]:
                    cnt += (tickets[k] - 1)
                else:
                    cnt += tickets[i]
        return cnt
```