LeetCode 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

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The translation of this question is to give an array, and each time the numbers in the array are -1 in sequence, and ask when the kth number can be reduced to 0.

I used brute force to write a while loop that sequentially changes a number to -1 each time, and finally ends when the kth number is 0, and the return value is the result.

class Solution {
    public int timeRequiredToBuy(int[] tickets, int k) {
        int i = 0;
        int result = 0;
        while (tickets[k] != 0) {
            if (tickets[i] > 0) {
                tickets[i]--;
                result++;
            }
            i = (i + 1) % tickets.length;
        }
        return result;
    }
}

If optimization is needed, we can decompose this array into two parts. One part is the number before k. They must be subtracted before k is subtracted, so for them, the number of times each number is subtracted is min(tickets[i], tickets[k]). The other part is the number after k, which can only be subtracted after subtracting k, so the number of times they are subtracted is min(tickets[k] - 1, tickets[i]). A for loop just loops through the array and adds them together.

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