Time Needed to Buy Tickets

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Arrays

There are n people in a line queuing to buy tickets, where the 0^th 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 i^th 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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the `tickets` array and the value of `k`?
Can the values in the `tickets` array be zero or negative?
Is `k` guaranteed to be a valid index within the `tickets` array bounds?
Could you provide a more concrete example to illustrate the process?
What should be returned if the `tickets` array is empty?

Brute Force Solution

Approach

We simulate the ticket-buying process round by round. We keep track of how much time passes as each person buys a ticket, one at a time, until the specified person buys their desired number of tickets. This approach mimics the actual real-world scenario.

Here's how the algorithm would work step-by-step:

Start with a time counter set to zero.
Look at the first person in line.
If they need to buy more than zero tickets, let them buy one ticket, increasing the time counter by one, and reduce their remaining ticket count by one.
If the first person doesn't need any more tickets (they have zero remaining), skip them.
Move to the next person in line and repeat the process.
Continue this process, going through each person in the line repeatedly, until the specific person we are interested in has bought all their desired tickets (their remaining ticket count is zero).
The final value of the time counter is the total time needed for the specified person to buy all their tickets.

Code Implementation

def time_needed_to_buy_tickets(tickets, person_to_target):
    time_elapsed = 0
    tickets_needed_by_target = tickets[person_to_target]

    while tickets_needed_by_target > 0:
        for current_person_index in range(len(tickets)):
            # Check if the current person still needs to buy tickets
            if tickets[current_person_index] > 0:
                tickets[current_person_index] -= 1
                time_elapsed += 1

                # Check if the current person is the target person

                if current_person_index == person_to_target:
                    tickets_needed_by_target -= 1
    return time_elapsed

Big(O) Analysis

Time Complexity

O(n*k) – The outer loop continues until tickets[k] becomes zero. In the worst case, tickets[k] might be equal to some value, let's call it 'k'. The inner loop iterates through all 'n' people in the tickets array in each round. Therefore, in the worst-case scenario, the outer loop could run up to 'k' times, while the inner loop always runs 'n' times. This results in a time complexity of approximately n * k operations, which is O(n*k), where k represents the initial value of tickets[k].

Space Complexity

O(1) – The provided solution operates directly on the input and uses only a few integer variables, like the time counter and potentially an index to iterate through the tickets array. No auxiliary data structures, such as lists, hash maps, or recursion stacks, are used. The space required by these variables remains constant regardless of the input size N (the number of people in the line). Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

We need to figure out how long it takes for a specific person to buy their tickets when people buy one ticket at a time in a circle. The trick is to simulate the buying process, but to stop as soon as the target person has bought all their tickets.

Here's how the algorithm would work step-by-step:

Start with the target person. We want to know when they're done buying tickets.
Imagine everyone is in a line to buy a ticket. Each person buys one ticket, and if they need more, they go back to the end of the line.
Keep track of how many tickets each person has left to buy.
Each time someone buys a ticket, reduce their remaining ticket count by one.
Continue this process, going through the line, until the target person has zero tickets left.
Count how many tickets were bought in total during this whole process. That total is the time it takes for the target person to finish buying their tickets.

Code Implementation

def time_needed_to_buy_tickets(tickets, target_person_index):
    total_time = 0
    tickets_needed = tickets[target_person_index]

    while tickets_needed > 0:
        for i in range(len(tickets)):
            # Stop when the target person has no more tickets to buy
            if tickets_needed == 0:
                break

            if tickets[i] > 0:
                tickets[i] -= 1
                total_time += 1

                # Check if the current person is the target person
                if i == target_person_index:
                    tickets_needed -= 1

    return total_time

Big(O) Analysis

Time Complexity

O(n*m) – The algorithm simulates the ticket buying process until the target person buys all their tickets. The outer loop implicitly iterates as long as the target person at tickets[k] has tickets to buy. The inner loop iterates through the entire tickets array (of size n) in each buying round. In the worst case, the target person tickets[k] has to buy 'm' number of tickets. Each person in front of tickets[k] has to buy almost 'm' number of tickets too. Thus the time complexity is O(n*m).

Space Complexity

O(1) – The algorithm simulates a queue and decrements the number of tickets for each person. It primarily uses integer variables to track the number of tickets each person needs and a counter for the total tickets bought. No auxiliary data structures, like arrays or hash maps, are created that scale with the number of people (N). Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty tickets array	Return 0 immediately since no tickets can be bought.
k is out of bounds (k < 0 or k >= len(tickets))	Raise an IndexError or return -1 to indicate an invalid input.
Single person in the queue (len(tickets) == 1)	Return tickets[0] since that's the number of turns it takes.
All people want the same number of tickets (tickets[i] == tickets[j] for all i, j)	The time taken is (k+1) * tickets[0] if tickets[0] or fewer are needed by each person; otherwise it's n * tickets[k], where n = len(tickets).
tickets[k] is 0	Return 0 immediately because the person at index k wants no tickets.
Large values in the tickets array (potential integer overflow)	Use a 64-bit integer type to store the result to prevent potential overflows.
k is the last person in the queue (k == len(tickets) - 1)	The algorithm should still proceed correctly, wrapping around to the beginning of the queue.
Tickets needed by other people are very small compared to tickets[k]	The algorithm should still proceed correctly and handle the case where k's tickets are depleted last.

Null or empty tickets array

How to Handle:

Return 0 immediately since no tickets can be bought.

k is out of bounds (k < 0 or k >= len(tickets))

How to Handle:

Raise an IndexError or return -1 to indicate an invalid input.

Single person in the queue (len(tickets) == 1)

How to Handle:

Return tickets[0] since that's the number of turns it takes.

All people want the same number of tickets (tickets[i] == tickets[j] for all i, j)

How to Handle:

The time taken is (k+1) * tickets[0] if tickets[0] or fewer are needed by each person; otherwise it's n * tickets[k], where n = len(tickets).

tickets[k] is 0

How to Handle:

Return 0 immediately because the person at index k wants no tickets.

Large values in the tickets array (potential integer overflow)

How to Handle:

Use a 64-bit integer type to store the result to prevent potential overflows.

k is the last person in the queue (k == len(tickets) - 1)

How to Handle:

The algorithm should still proceed correctly, wrapping around to the beginning of the queue.

Tickets needed by other people are very small compared to tickets[k]

How to Handle:

The algorithm should still proceed correctly and handle the case where k's tickets are depleted last.