> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Time Needed to Buy Tickets Python Solution

> Tested Python solution for LeetCode 2073 with 18 pytest cases. Generate a practice environment with lcpy.

LeetCode 2073, [Easy](/catalog/easy). Topics: [Array](/catalog/topics/array), [Queue](/catalog/topics/queue), [Simulation](/catalog/topics/simulation). [View on LeetCode](https://leetcode.com/problems/time-needed-to-buy-tickets/description/).

Generate this problem as a practice environment: tested reference solution, 18 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 2073   # by problem number
lcpy gen -s time_needed_to_buy_tickets   # by problem name
```

## Problem

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 initially at position k (0-indexed) to finish buying tickets.

### Examples

```
Input: tickets = [2,3,2], k = 2
Output: 6
```

**Explanation:**

* The queue starts as \[2,3,2], where the kth person is underlined.
* After the person at the front has bought a ticket, the queue becomes \[3,2,1] at 1 second.
* Continuing this process, the queue becomes \[2,1,2] at 2 seconds.
* Continuing this process, the queue becomes \[1,2,1] at 3 seconds.
* Continuing this process, the queue becomes \[2,1] at 4 seconds. Note: the person at the front left the queue.
* Continuing this process, the queue becomes \[1,1] at 5 seconds.
* Continuing this process, the queue becomes \[1] at 6 seconds. The kth person has bought all their tickets, so return 6.

```
Input: tickets = [5,1,1,1], k = 0
Output: 8
```

**Explanation:**

* The queue starts as \[5,1,1,1], where the kth person is underlined.
* After the person at the front has bought a ticket, the queue becomes \[1,1,1,4] at 1 second.
* Continuing this process for 3 seconds, the queue becomes \[4] at 4 seconds.
* Continuing this process for 4 seconds, the queue becomes \[] at 8 seconds. The kth person has bought all their tickets, so return 8.

### Constraints

* n == tickets.length
* 1 \<= n \<= 100
* 1 \<= tickets\[i] \<= 100
* 0 \<= k \< n

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/time_needed_to_buy_tickets/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/time_needed_to_buy_tickets/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    # Time: O(n)
    # Space: O(1)
    def time_required_to_buy(self, tickets: list[int], k: int) -> int:
        target = tickets[k]
        total = 0
        for i, need in enumerate(tickets):
            if i <= k:
                total += min(need, target)
            else:
                total += min(need, target - 1)
        return total
```

## Complexity

| Time | Space |
| - | - |
| O(n) | O(1) |

## Tags

[NeetCode All](/catalog/neetcode).


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