# Write a Program to Calculate Time Needed to Buy Tickets

# 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 ```

/** * @param {number[]} tickets * @param {number} k * @return {number} */ var timeRequiredToBuy = function(tickets, k) { let out = 0; for (let i = 0; i < tickets.length; i++) { if (i === k || tickets[i] < tickets[k]) { out += tickets[i]; } else if (i > k) { out += tickets[k] - 1; } else { out += tickets[k]; } } return out; };