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;
                }
            }
        }
    };
    
  • // Todo
    
  • # 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))
                    
            
            
    
    

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;
        }
    };
    
  • // Todo
    
  • # 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))
                    
            
            
    
    

Discuss

https://leetcode.com/problems/time-needed-to-buy-tickets/discuss/1576932/C%2B%2B-One-Pass

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