LeetCode-in-Java
2073. Time Needed to Buy Tickets
Easy
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.length1 <= n <= 1001 <= tickets[i] <= 1000 <= k < n
Solution
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;
}
}