Time Needed to Buy Tickets

easy array queue simulation

Problem

Each step the first person buys one ticket then goes to the back unless done. Return the time (in steps) until person at index k buys all their tickets.

Inputtickets = [2, 3, 2], k = 2

Output6

Ahead (idx < k) buy at most t[k]; behind at most t[k]−1.

tickets (comma-separated)

def time_required_to_buy(tickets, k):
    total = 0
    for i, t in enumerate(tickets):
        if i <= k:
            total += min(t, tickets[k])
        else:
            total += min(t, tickets[k] - 1)
    return total

function timeRequiredToBuy(tickets, k) {
  let total = 0;
  for (let i = 0; i < tickets.length; i++) {
    if (i <= k) total += Math.min(tickets[i], tickets[k]);
    else total += Math.min(tickets[i], tickets[k] - 1);
  }
  return total;
}

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

int timeRequiredToBuy(vector<int>& tickets, int k) {
    int total = 0;
    for (int i = 0; i < (int)tickets.size(); i++) {
        if (i <= k) total += min(tickets[i], tickets[k]);
        else total += min(tickets[i], tickets[k] - 1);
    }
    return total;
}

Explanation

You could literally simulate the line — buy a ticket, send the person to the back, repeat — but there is a much faster counting shortcut that needs just one pass and no queue at all.

The key insight is that person k stops the instant they buy their last ticket, after tickets[k] rounds. So we ask, for everyone else, how many tickets they manage to buy before that moment.

Anyone at or ahead of k (index i <= k) gets a turn in every one of those rounds, so they buy min(tickets[i], tickets[k]). Anyone behind k (index i > k) gets one fewer chance, because the line stops right when k finishes — so they buy min(tickets[i], tickets[k] - 1).

Each ticket bought is one unit of time, so we just add up all these contributions with total += ... and return the sum.

Example: tickets = [2, 3, 2], k = 2. Person 0 (ahead): min(2, 2) = 2. Person 1 (ahead): min(3, 2) = 2. Person 2 (k itself): min(2, 2) = 2. Total = 6, all in O(n).

Time: O(n) Space: O(1)