Formatted question description: https://leetcode.ca/all/2073.html

# 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.length
• 1 <= n <= 100
• 1 <= tickets[i] <= 100
• 0 <= k < n

Similar Questions:

## Solution 1. Brute Force Simulation

// OJ: https://leetcode.com/problems/time-needed-to-buy-tickets/
// Time: O(SUM(A))
// Space: O(1)
class Solution {
public:
int timeRequiredToBuy(vector<int>& A, int k) {
int step = 0;
while (true) {
for (int i = 0; i < A.size(); ++i) {
if (A[i] == 0) continue;
A[i]--;
++step;
if (A[k] == 0) return step;
}
}
}
};


## Solution 2. One Pass

For i <= k, A[i] contributes min(A[k], A[i]) steps.

For i > k, A[i] contributes min(A[k] - 1, A[i]) steps.

// OJ: https://leetcode.com/problems/time-needed-to-buy-tickets/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int timeRequiredToBuy(vector<int>& A, int k) {
int ans = 0;
for (int i = 0; i < A.size(); ++i) {
ans += min(A[k] - (i > k), A[i]);
}
return ans;
}
};