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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