2073. Time Needed to Buy Tickets

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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 {

/**
* @param Integer[] $tickets
* @param Integer $k
* @return Integer
*/
function timeRequiredToBuy($tickets, $k) {
$ans = 0;

foreach ($tickets as $i => $ticket) {
if ($i <= $k) {
$ans += min($ticket, $tickets[$k]);
} else {
$ans += min($ticket, $tickets[$k] – 1);
}
}

return $ans;
}
}

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