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2073. Time Needed to Buy Tickets
Description
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
Solutions

class Solution { public int timeRequiredToBuy(int[] tickets, int k) { int ans = 0; for (int i = 0; i < tickets.length; i++) { if (i <= k) { ans += Math.min(tickets[k], tickets[i]); } else { ans += Math.min(tickets[k]  1, tickets[i]); } } return ans; } }

class Solution { public: int timeRequiredToBuy(vector<int>& tickets, int k) { int ans = 0; for (int i = 0; i < tickets.size(); ++i) { if (i <= k) { ans += min(tickets[k], tickets[i]); } else { ans += min(tickets[k]  1, tickets[i]); } } return ans; } };

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

func timeRequiredToBuy(tickets []int, k int) int { ans := 0 for i, t := range tickets { if i <= k { ans += min(tickets[k], t) } else { ans += min(tickets[k]1, t) } } return ans }

function timeRequiredToBuy(tickets: number[], k: number): number { const n = tickets.length; let target = tickets[k]  1; let ans = 0; // round1 for (let i = 0; i < n; i++) { let num = tickets[i]; if (num <= target) { ans += num; tickets[i] = 0; } else { ans += target; tickets[i] = target; } } // round2 for (let i = 0; i <= k; i++) { let num = tickets[i]; ans += num > 0 ? 1 : 0; } return ans; }