2073. Time Needed to Buy Tickets

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`

2073. Time Needed to Buy Tickets
``````struct Solution;

impl Solution {
fn time_required_to_buy(tickets: Vec<i32>, k: i32) -> i32 {
let mut res = 0;
let n = tickets.len();
let mut arr = vec![0; n];
let mut i = 0;
let k = k as usize;
while arr[k] != tickets[k] {
if arr[i] < tickets[i] {
arr[i] += 1;
res += 1;
}

i += 1;
i %= n;
}
res
}
}

#[test]
fn test() {
let tickets = vec![2, 3, 2];
let k = 2;
let res = 6;