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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2001_2100.s2073_time_needed_to_buy_tickets;

// #Easy #Array #Simulation #Queue #2022_05_27_Time_0_ms_(100.00%)_Space_41.9_MB_(45.92%)

/**
 * 2073 - Time Needed to Buy Tickets\.
 *
 * Easy
 *
 * There are `n` people in a line queuing to buy tickets, where the 0^th 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 i^th 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`
**/
public 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;
    }
}