[LeetCode] Time Needed to Buy Tickets

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.

1
2
3
4
5
6
7
8
9
10
class Solution {
public:
int timeRequiredToBuy(vector<int>& A, int k) {
int res = 0;
for(int i = 0; i < A.size(); i++) {
res += min(A[i], A[k] - (bool)(i > k));
}
return res;
}
};
Author: Song Hayoung
Link: https://songhayoung.github.io/2024/04/09/PS/LeetCode/time-needed-to-buy-tickets/
Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.