Time Needed to Buy Tickets

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class Solution {
public:
    int timeRequiredToBuy(vector<int>& A, int k) {
        int ans = 0;
        for (int i = 0; i < A.size(); ++i) {
            ans += min(A[k] - (i > k), A[i]);
        }
        return ans;
    }
};

The C++ function, timeRequiredToBuy, calculates the time required to buy all the items in a vector A, given that the k-th item is bought first. The time required to buy an item is given by its value in A.

The function works as follows:

It initializes a variable ans to 0. This will hold the total time required to buy all the items.
It then iterates over all the items in A. For each item:
- If the index of the item i is less than or equal to k, the time required to buy the item is just its value in A.
- If the index of the item i is greater than k, the time required to buy the item is the minimum of its value in A and the value of the k-th item in A minus i - k. This is because we’ve already spent i - k units of time buying previous items, and we can’t spend more time buying this item than the value of the k-th item.
The function adds the time required to buy each item to ans.
Finally, the function returns ans, which is the total time required to buy all the items.

For example, if A = {2, 3, 1, 2, 4} and k = 2, the function will return 2 + 3 + 1 + 1 + 2 = 9.

This function assumes that k is a valid index in A, and that all values in A are non-negative. It also assumes that the time required to buy an item can’t be more than the value of the k-th item. If these assumptions are not met, the function might not work correctly.

Python solution:

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class Solution:
    def timeRequiredToBuy(self, tickets, k):
        ans = 0
        for i in range(len(tickets)):
            ans += min(tickets[k] - (i > k), tickets[i])
        return ans

In this function, tickets is the list of integers representing the number of tickets each person wants to buy, and k is the position of the person you’re focused on. For each person in the line (from position 0 to len(tickets)-1), we calculate how many tickets they’ll be able to buy before the person at position k has bought all of theirs. We add these numbers together to get the total number of tickets sold before the person at position k has bought all of theirs. The time required for the person at position k to buy all their tickets is equal to this total.

We are looking at a line of people waiting to buy tickets. Each person has a certain number of tickets they want to buy, and it takes 1 second to buy each ticket. After a person buys a ticket, if they want to buy more, they go to the back of the line.

The value of k tells us the position of a person in line we are interested in.

Elements before k (and k itself): Each of these people will get to buy tickets as many times as the smaller of these two numbers: the number of tickets the k-th person wants to buy (t[k]), or the number of tickets they want to buy (t[i]). This is because everyone keeps going to the back of the line after they buy a ticket, so everyone before (and including) the k-th person will get to buy tickets until either they have all the tickets they want or the k-th person has all the tickets they want.
Elements after k: Each of these people will get to buy tickets as many times as the smaller of these two numbers: one less than the number of tickets the k-th person wants to buy (t[k] - 1), or the number of tickets they want to buy (t[i]). This is because they start buying tickets after the k-th person has bought their first ticket, so they keep buying tickets until either they have all the tickets they want or the k-th person has all but one of the tickets they want.

The number of tickets each person gets to buy depends on their position relative to the k-th person and on how many tickets the k-th person wants to buy.

Imagine we have 4 people (A, B, C, D) in line to buy tickets and you are person A. Each person needs a different number of tickets: A needs 4, B needs 7, C needs 3, D needs 4.

We’re trying to figure out how long each person has to wait to get all their tickets.

For person A, you don’t have to wait for anyone since you’re first in line. So, it takes you 4 seconds to get all your tickets.
For person B, they have to wait for you (A) to get all 4 tickets first. Then, they can buy their own tickets. So it’s 4 (waiting for A) + 3 (buying their own) = 7 seconds.
For person C, they have to wait for you (A) and B to get all their tickets. Because C needs only 3 tickets, they’ll get all their tickets by the time you and B get your tickets. So, it’s 4 (waiting for A) + 3 (waiting for B) + 3 (buying their own) = 10 seconds.
For person D, they also have to wait for A, B, and C. But here’s the catch: D wants 4 tickets, but they can only get 3 of them before you (A) have gotten all of yours. So D gets the last ticket after you’re done. So, it’s 10 (waiting for A, B, C) + 3 (buying their own) = 13 seconds.

So the rule is, for all people who come after you (A):

If they need fewer tickets than you, they just wait for you and the people before them and buy all their tickets.
If they need the same number or more tickets than you, they can buy all but one of their tickets while waiting for you and the people before them, and get the last ticket after you’re done.

That’s the pattern!

Let’s represent this scenario in a table. We have four people: A, B, C, D. They need 4, 7, 3, 4 tickets respectively.

	Queue Position	Tickets Needed	Wait Time	Own Tickets Time	Total Time
A	1st	4	0	4	4
B	2nd	7	4	3	7
C	3rd	3	7	3	10
D	4th	4	10	3	13

The Queue Position column indicates where each person is in line.
The Tickets Needed column shows how many tickets each person wants to buy.
The Wait Time column calculates how long each person has to wait until it’s their turn. They’re essentially waiting for everyone in front of them to get their tickets.
The Own Tickets Time column shows how many tickets they are able to buy while waiting for others. If they need the same or more tickets than the person in front of them (like person D), they will have to wait for one extra ticket.
The Total Time column is the sum of Wait Time and Own Tickets Time.

From this table, you can see that if a person needs fewer or same tickets than the person in front of them (like person B and C), they can buy all their tickets while waiting. If they need more tickets than the person in front of them (like person D), they will have to wait for one extra ticket.

Let’s illustrate this scenario in a table format. Here, A, B, C, and D are the people and the numbers represent the tickets they need:

Round	A	B	C	D
1	4	7	3	4
2	3	6	2	3
3	2	5	1	2
4	1	4	0	1
5	0	NA	NA	NA

In each round, all people in the queue buy one ticket. As you can see, by the end of the 4th round, person C has bought all of their tickets and leaves the queue. By the end of the 5th round, person A has also bought all of their tickets and leaves the queue. Persons B and D will continue to buy tickets until they have all they need.

The person’s time is calculated as their wait time plus the time they spend buying their own tickets. The wait time is equal to the tickets the person in front of them needs, while their own tickets time is the number of tickets they need minus one (if their need is more than or equal to the person in front of them) or is the number of tickets they need (if their need is less than the person in front of them).

The essence of this problem is about understanding the order and the rounds.

When you’re at position k, everyone in front of you has to buy their tickets before you can start buying yours. That’s why we take the minimum between a[i] and a[k] for all the people ahead of you (from position 0 to k-1). Here, a[i] represents the tickets the i-th person needs, and a[k] is the tickets you need. If they need more tickets than you, they will still be in line after you start buying your tickets. However, if they need fewer tickets than you, they will leave before you finish.

For the buyers after you (from position k+1 to n-1), when it’s your turn to buy tickets, you will buy your first ticket and then go back to the end of the line. That’s why we use a[k] - 1 for them. If they need fewer tickets than you, they will leave the line before you get your last ticket. But if they need more tickets or the same number of tickets as you, they will still be in line when you’re buying your last ticket.

The sum of these wait times plus the time you spend buying your own tickets (which is a[k] times) will be the total time required for the person at position k to finish buying tickets.

To make this clear, let’s run through the example you provided: tickets = {3, 4, 5, 2, 8} with k=2.

At the beginning, the ticket line looks like this:

[3, 4, 5, 2, 8]

After the first round, each person has bought a ticket and the line becomes:

[2, 3, 4, 1, 7]

After the second round, it changes to:

[1, 2, 3, 0, 6]

And so on, until we reach the state where the person at position k=2 has bought all their tickets, resulting in the line:

[0, 0, 0, 0, 4]

So, to summarize, when k (the person in question) finishes buying their tickets, anyone in line who initially wanted to buy fewer or equal number of tickets than k will have also finished. Therefore, when you’re counting time for people at positions k+1 to n-1 (after k), you’re considering the fact that when k starts buying tickets, they will buy their first one and then go back to the end of the line. Hence, you check for min(a[k] - 1, a[i]) times. If they need fewer tickets than a[k] - 1, they will leave the line before k gets their last ticket. But if they need more tickets or the same number as a[k] - 1, they will still be in line when k is buying their last ticket.

Time taken by people before k: Here we iterate through each person before k. Either this person gets all their tickets and leaves, or person k gets all their tickets and the time for this person stops counting. This is represented by the term min(tickets[k], tickets[i]).
Time taken by person k: This is the time for person k to get all of their tickets, which is simply tickets[k].
Time taken by people after k: For the people who are after k in the line, we stop counting time for them once k has only 1 ticket left to buy, because at that point k is at the end of the line. So, either the person gets all their tickets and leaves, or k has just 1 last ticket left. This is represented by the term min(tickets[k] - 1, tickets[i]).

In each of these three cases, we are determining how many tickets a person in line will be able to buy before person k has bought all their tickets. The total time is the sum of these three parts.

This problem is about visualizing and simulating the queue of people buying tickets.

The key insight, as you correctly pointed out, is understanding that when the k-th person finishes buying all their tickets, the people behind them in the queue who need more tickets than the k-th person can only have bought (k-1) tickets at that point. This is because once the k-th person is done, they’re not in the queue anymore, so the people behind them don’t get the chance to buy that k-th ticket.

This insight is what leads to the min(tickets[k] - 1, tickets[i]) part of the solution, ensuring we correctly calculate the time for people after k in the line. The total time is then the sum of the time taken by the people before k, the k-th person themselves, and the people after k, as per the logic we discussed earlier.

Understanding these queue dynamics and how each person’s ticket buying affects others is crucial to solving this problem. You’ve captured this concept excellently.

just use queue data structure ! and push the indexes in it. and increase the time count after every iteration.

Yes, you’re right! A queue data structure would indeed be a good fit for solving this problem due to the nature of the ticket buying process.

Here’s a rough idea of how you might implement this:

Initialize an empty queue, then populate it with the indices of people in line (from 0 to n-1).
Set a variable time to 0, which will keep track of the total time passed.
Then, perform a while loop that continues as long as the queue is not empty:
- Dequeue an index from the front of the queue.
- If the person at this index still needs to buy tickets (i.e., tickets[index] > 0), decrease the ticket count for this person by 1, and if they still need more tickets (i.e., tickets[index] > 0), enqueue their index back into the queue.
- Increase the time count by 1.
Stop when you’ve served the person at position k for the last time (i.e., when they no longer need to buy tickets).

This approach maintains the order of the queue while ensuring that each person goes to the back of the line after buying a ticket, mirroring the process described in the problem.

Here’s a Python solution using a queue implemented with a list:

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from typing import List

class Solution:
    def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
        queue = [(ticket, idx) for idx, ticket in enumerate(tickets)]
        time = 0

        while True:
            ticket, idx = queue.pop(0)
            ticket -= 1
            time += 1

            if idx == k and ticket == 0:
                break
            
            if ticket > 0:
                queue.append((ticket, idx))

        return time

In this solution, we start by initializing the queue with pairs of ticket numbers and their respective indices. We then enter a loop where we serve each person (decrement their ticket count), increment the time, and if they still need tickets, put them back in the queue. We continue this until we’ve served the person at position k for the last time.

This problem is basically a real-world queue simulation with time element in it.

Yes, you’re absolutely correct. This problem is a simulation of a queue where each person takes a certain amount of time to complete their task (in this case, buying tickets). By simulating the queue in the program, we can determine exactly how much time it would take for the person at the kth position to finish buying their tickets.

In the queue, each person takes exactly 1 second to buy a ticket. If a person still needs to buy more tickets, they are put back into the queue to wait their turn again. The person at the kth position may need to go through the queue multiple times to buy all of their tickets, and the total time it takes them to do so is the solution to the problem.

This kind of problem, where you need to simulate a real-world scenario in order to find the solution, can be quite common in coding challenges and interviews, as it tests your ability to translate real-world logic into code.