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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 0^{th}
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 0indexed integer array tickets
of length n
where the number of tickets that the i^{th}
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
(0indexed) 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/timeneededtobuytickets/ // 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; } } } };

// Todo

class Solution: def timeRequiredToBuy(self, tickets: List[int], k: int) > int: ans = 0 for i, t in enumerate(tickets): if i <= k: ans += min(tickets[k], t) else: ans += min(tickets[k]  1, t) return ans ############ # 2073. Time Needed to Buy Tickets # https://leetcode.com/problems/timeneededtobuytickets class Solution: def timeRequiredToBuy(self, tickets: List[int], k: int) > int: if tickets[k] == 0: return 0 A = deque([(index, t) for index, t in enumerate(tickets) if t > 0]) time = 0 while A: n = len(A) for _ in range(n): index, t = A.popleft() time += 1 if index == k and t == 1: return time if t  1 > 0: A.append((index, t  1))
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/timeneededtobuytickets/ // 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; } };

// Todo

class Solution: def timeRequiredToBuy(self, tickets: List[int], k: int) > int: ans = 0 for i, t in enumerate(tickets): if i <= k: ans += min(tickets[k], t) else: ans += min(tickets[k]  1, t) return ans ############ # 2073. Time Needed to Buy Tickets # https://leetcode.com/problems/timeneededtobuytickets class Solution: def timeRequiredToBuy(self, tickets: List[int], k: int) > int: if tickets[k] == 0: return 0 A = deque([(index, t) for index, t in enumerate(tickets) if t > 0]) time = 0 while A: n = len(A) for _ in range(n): index, t = A.popleft() time += 1 if index == k and t == 1: return time if t  1 > 0: A.append((index, t  1))
Discuss
https://leetcode.com/problems/timeneededtobuytickets/discuss/1576932/C%2B%2BOnePass