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2073. Time Needed to Buy Tickets

Description

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 initially at position k (0-indexed) to finish buying tickets.

 

Example 1:

Input: tickets = [2,3,2], k = 2

Output: 6

Explanation:

  • The queue starts as [2,3,2], where the kth person is underlined.
  • After the person at the front has bought a ticket, the queue becomes [3,2,1] at 1 second.
  • Continuing this process, the queue becomes [2,1,2] at 2 seconds.
  • Continuing this process, the queue becomes [1,2,1] at 3 seconds.
  • Continuing this process, the queue becomes [2,1] at 4 seconds. Note: the person at the front left the queue.
  • Continuing this process, the queue becomes [1,1] at 5 seconds.
  • Continuing this process, the queue becomes [1] at 6 seconds. The kth person has bought all their tickets, so return 6.

Example 2:

Input: tickets = [5,1,1,1], k = 0

Output: 8

Explanation:

  • The queue starts as [5,1,1,1], where the kth person is underlined.
  • After the person at the front has bought a ticket, the queue becomes [1,1,1,4] at 1 second.
  • Continuing this process for 3 seconds, the queue becomes [4] at 4 seconds.
  • Continuing this process for 4 seconds, the queue becomes [] at 8 seconds. The kth person has bought all their tickets, so return 8.

 

Constraints:

  • n == tickets.length
  • 1 <= n <= 100
  • 1 <= tickets[i] <= 100
  • 0 <= k < n

Solutions

Solution 1: Single Pass

According to the problem description, when the $k^{th}$ person finishes buying tickets, all the people in front of the $k^{th}$ person will not buy more tickets than the $k^{th}$ person, and all the people behind the $k^{th}$ person will not buy more tickets than the $k^{th}$ person minus $1$.

Therefore, we can traverse the entire queue. For the $i^{th}$ person, if $i \leq k$, the time to buy tickets is $\min(\textit{tickets}[i], \textit{tickets}[k])$; otherwise, the time to buy tickets is $\min(\textit{tickets}[i], \textit{tickets}[k] - 1)$. We sum the buying time for all people to get the result.

The time complexity is $O(n)$, where $n$ is the length of the queue. The space complexity is $O(1)$.

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class Solution:
    def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
        ans = 0
        for i, x in enumerate(tickets):
            ans += min(x, tickets[k] if i <= k else tickets[k] - 1)
        return ans

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

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

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func timeRequiredToBuy(tickets []int, k int) (ans int) {
    for i, x := range tickets {
        t := tickets[k]
        if i > k {
            t--
        }
        ans += min(x, t)
    }
    return
}

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function timeRequiredToBuy(tickets: number[], k: number): number {
    let ans = 0;
    const n = tickets.length;
    for (let i = 0; i < n; ++i) {
        ans += Math.min(tickets[i], i <= k ? tickets[k] : tickets[k] - 1);
    }
    return ans;
}

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