Question

2073. Time Needed to Buy Tickets

中文文档

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 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

Solutions

Solution 1

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

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;
  }
};

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;
}