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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这道题翻译过来就是给了一个数组，每次按顺序把数组里的数字-1，问第k个数啥时候能减到0。

我就brute force写了个while loop每次按顺序把一个数字-1，最后在第k个数为0的时候结束，return的值就是结果了。

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;
    }
}

如果需要优化的话，我们可以把这个数组分解成两个部分。一个部分是在k之前的数，在减k之前肯定得先减它们，于是对它们来说，每个数字被减的次数是min(tickets[i], tickets[k])。另一部分是在k之后的数，只能在减k之后减它们，于是它们被减的次数是min(tickets[k] - 1, tickets[i])。一个for loop循环遍历数组以后相加就行。

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;
    }
}