Formatted question description: https://leetcode.ca/all/2073.html

# 2073. Time Needed to Buy Tickets (Easy)

There are `n`

people in a line queuing to buy tickets, where the `0`

person is at the ^{th}**front** of the line and the `(n - 1)`

person is at the ^{th}**back** of the line.

You are given a **0-indexed** integer array `tickets`

of length `n`

where the number of tickets that the `i`

person would like to buy is ^{th}`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 = 2Output:6Explanation:- 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 = 0Output:8Explanation:- 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`

**Similar Questions**:

## Solution 1. Brute Force Simulation

```
// OJ: https://leetcode.com/problems/time-needed-to-buy-tickets/
// Time: O(SUM(A))
// Space: O(1)
class Solution {
public:
int timeRequiredToBuy(vector<int>& A, int k) {
int step = 0;
while (true) {
for (int i = 0; i < A.size(); ++i) {
if (A[i] == 0) continue;
A[i]--;
++step;
if (A[k] == 0) return step;
}
}
}
};
```

## Solution 2. One Pass

For `i <= k`

, `A[i]`

contributes `min(A[k], A[i])`

steps.

For `i > k`

, `A[i]`

contributes `min(A[k] - 1, A[i])`

steps.

```
// OJ: https://leetcode.com/problems/time-needed-to-buy-tickets/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int timeRequiredToBuy(vector<int>& A, int k) {
int ans = 0;
for (int i = 0; i < A.size(); ++i) {
ans += min(A[k] - (i > k), A[i]);
}
return ans;
}
};
```

## Discuss

https://leetcode.com/problems/time-needed-to-buy-tickets/discuss/1576932/C%2B%2B-One-Pass