2073. Time Needed to Buy Tickets

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 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

Solution (Java)

class Solution {
    public int timeRequiredToBuy(int[] tickets, int k) {
        int res = 0;
        for (int i = 0; i < tickets.length; i++) {
            if (i <= k) {
                res += Math.min(tickets[k], tickets[i]);
            } else {
                res += Math.min(tickets[k] - 1, tickets[i]);
            }
        }
        return res;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).