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`

**Solution 1: Simulation**

Time complexity: O(n * tickets[k])

Space complexity: O(n) / O(1)

## C++

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// Author: huahua class Solution { public: int timeRequiredToBuy(vector<int>& tickets, int k) { const int n = tickets.size(); queue<pair<int, int>> q; for (int i = 0; i < n; ++i) q.emplace(i, tickets[i]); int ans = 0; while (!q.empty()) { ++ans; auto [i, t] = q.front(); q.pop(); if (--t == 0 && k == i) return ans; if (t) q.emplace(i, t); } return -1; } }; |

**Solution 2: Math**

Each person before k will have tickets[k] rounds, each person after k will have tickets[k] – 1 rounds.

Time complexity: O(n)

Space complexity: O(1)

## C++

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// Author: huahua class Solution { public: int timeRequiredToBuy(vector<int>& tickets, int k) { int ans = 0; for (int i = 0; i < tickets.size(); ++i) ans += min(tickets[i], tickets[k] - (i > k)); return ans; } }; |