LeetCode in Kotlin

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

Solution

class Solution {
    fun timeRequiredToBuy(tickets: IntArray, k: Int): Int {
        var res = 0
        for (i in tickets.indices) {
            res += if (i <= k) {
                tickets[k].coerceAtMost(tickets[i])
            } else {
                (tickets[k] - 1).coerceAtMost(tickets[i])
            }
        }
        return res
    }
}

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