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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 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
Similar Questions:
Solution 1. Brute Force Simulation
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// 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; } } } }; -
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/time-needed-to-buy-tickets 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)) -
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; } } -
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 } func min(a, b int) int { if a < b { return a } return b } -
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; }
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.
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// 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; } }; -
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/time-needed-to-buy-tickets 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)) -
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; } } -
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 } func min(a, b int) int { if a < b { return a } return b } -
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; }
Discuss
https://leetcode.com/problems/time-needed-to-buy-tickets/discuss/1576932/C%2B%2B-One-Pass