LeetCode 2073. Time Needed to Buy Tickets Solution in Java, C++, Python & More | Explanation + Code

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 ≤ k, the time to buy tickets is min(tickets[i], tickets[k]); otherwise, the time to buy tickets is min(tickets[i], 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).

PythonJavaC++GoTypeScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)