Problem Description

You have n people standing in a line waiting to buy tickets. Each person is numbered from 0 to n-1, where person 0 is at the front of the line and person n-1 is at the back.

You're given an array tickets where tickets[i] represents how many tickets the ith person wants to buy.

The ticket buying process works as follows:

• Each person takes exactly 1 second to buy 1 ticket
• A person can only buy 1 ticket at a time
• After buying a ticket, if they need more tickets, they must go to the back of the line instantly
• When a person has bought all their tickets, they leave the line

Your task is to find how many seconds it takes for the person at position k to finish buying all their tickets.

Example walkthrough:

If tickets = [2, 3, 2] and k = 2:

• Initially: Person 0 needs 2 tickets, Person 1 needs 3 tickets, Person 2 needs 2 tickets
• Second 1: Person 0 buys 1 ticket, goes to back (now needs 1 more)
• Second 2: Person 1 buys 1 ticket, goes to back (now needs 2 more)
• Second 3: Person 2 buys 1 ticket, goes to back (now needs 1 more)
• Second 4: Person 0 buys their last ticket and leaves
• Second 5: Person 1 buys 1 ticket, goes to back (now needs 1 more)
• Second 6: Person 2 buys their last ticket and leaves

Person 2 finishes at second 6, so the answer is 6.

The key insight is that when person k finishes buying tickets:

• People before position k in the original line will buy at most tickets[k] tickets
• People after position k in the original line will buy at most tickets[k] - 1 tickets (since they get fewer chances before person k finishes)

Intuition

Instead of simulating the entire queue process, let's think about what happens to each person when person k finishes buying their tickets.

The key observation is that we don't need to track the queue's state at each second. We just need to count how many tickets each person will have bought by the time person k finishes.

Let's think about two scenarios:

1. People originally in front of or at position k (positions 0 to k): These people get to buy tickets before person k in every round of the queue. So if person k needs tickets[k] tickets, these people will have the opportunity to buy up to tickets[k] tickets. However, if someone needs fewer tickets than tickets[k], they'll leave the line early after buying all they need. Therefore, each person at position i where i <= k contributes min(tickets[i], tickets[k]) seconds.

2. People originally behind position k (positions k+1 to n-1): These people start behind person k, so in the final round when person k buys their last ticket, these people won't get another chance. They get one less opportunity compared to person k. Therefore, each person at position i where i > k contributes min(tickets[i], tickets[k] - 1) seconds.

The total time is simply the sum of all these individual contributions. Each ticket bought by anyone before person k finishes adds 1 second to our answer.

This transforms the problem from a complex queue simulation into a simple calculation: for each person, determine how many tickets they'll buy before person k finishes, and sum these amounts.

Learn more about Queue patterns.

Solution Approach

The implementation follows directly from our intuition using a single pass through the array:

1. Initialize a counter: Start with ans = 0 to accumulate the total time.

2. Iterate through each person: Use enumerate() to get both the index i and the number of tickets x that person i wants.

3. Calculate contribution for each person:

   • If i <= k (person is at or before position k): They can buy up to tickets[k] tickets, but limited by their own need. So we add min(x, tickets[k]) to the answer.
   • If i > k (person is after position k): They get one less opportunity, so we add min(x, tickets[k] - 1) to the answer.

4. Return the total: The accumulated sum represents the total seconds needed.

The solution can be written concisely using a ternary operator:

ans += min(x, tickets[k] if i <= k else tickets[k] - 1)

This expression elegantly handles both cases:

• When i <= k, we compare with tickets[k]
• When i > k, we compare with tickets[k] - 1

Time Complexity: O(n) where n is the number of people, as we make a single pass through the array.

Space Complexity: O(1) as we only use a constant amount of extra space for the counter variable.

The beauty of this solution is that it avoids simulating the queue operations entirely, reducing what could be an O(n * max(tickets)) simulation to a simple O(n) calculation.

Example Walkthrough

Let's walk through a concrete example with tickets = [5, 1, 1, 1] and k = 0.

We want to find how long it takes for person 0 (who needs 5 tickets) to finish buying all their tickets.

Using our solution approach:

Instead of simulating the queue, we calculate how many tickets each person buys before person 0 finishes:

1. Person 0 (i = 0, needs 5 tickets):

   • Since i = 0 <= k = 0, this person is at or before position k
   • They will buy min(5, tickets[0]) = min(5, 5) = 5 tickets
   • Contribution: 5 seconds

2. Person 1 (i = 1, needs 1 ticket):

   • Since i = 1 > k = 0, this person is after position k
   • They get one less opportunity than person 0
   • They will buy min(1, tickets[0] - 1) = min(1, 4) = 1 ticket
   • Contribution: 1 second

3. Person 2 (i = 2, needs 1 ticket):

   • Since i = 2 > k = 0, this person is after position k
   • They will buy min(1, tickets[0] - 1) = min(1, 4) = 1 ticket
   • Contribution: 1 second

4. Person 3 (i = 3, needs 1 ticket):

   • Since i = 3 > k = 0, this person is after position k
   • They will buy min(1, tickets[0] - 1) = min(1, 4) = 1 ticket
   • Contribution: 1 second

Total time = 5 + 1 + 1 + 1 = 8 seconds

This makes sense because:

• Person 0 buys 5 tickets total (taking 5 seconds for their purchases)
• Each of the 3 people behind person 0 gets to buy 1 ticket before person 0 completes (adding 3 more seconds)
• The people behind person 0 can't buy more than 4 tickets each (tickets[0] - 1), but they only need 1 ticket anyway

The key insight is that we don't need to track the queue state at each second - we just count how many tickets everyone buys before person k finishes.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the tickets array. The algorithm iterates through the tickets array exactly once using a single for loop with enumerate. Within each iteration, it performs constant time operations: accessing array elements, comparison with min(), and addition. Therefore, the overall time complexity is linear with respect to the input size.

The space complexity is O(1). The algorithm only uses a fixed amount of extra space regardless of the input size. It maintains only two variables: ans for accumulating the result and the loop variables i and x from enumerate. No additional data structures that scale with the input are created, resulting in constant space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Off-by-One Error for People After Position k

The Mistake: A common error is treating all people the same way, forgetting that people after position k get one fewer opportunity to buy tickets before person k finishes.

# INCORRECT - treats everyone the same
for i, x in enumerate(tickets):
    total_time += min(x, tickets[k])  # Wrong for i > k!

Why It Happens: It's intuitive to think everyone gets the same number of chances, but in the circular queue, person k exits immediately after buying their last ticket, giving people behind them one less turn.

The Fix: Always distinguish between people before/at position k and those after:

if index <= k:
    total_time += min(ticket_count, tickets[k])
else:
    total_time += min(ticket_count, tickets[k] - 1)

Pitfall 2: Simulating the Queue Instead of Calculating Directly

The Mistake: Implementing an actual queue simulation with dequeue and enqueue operations:

# INEFFICIENT - simulating the actual queue
from collections import deque
queue = deque([(i, tickets[i]) for i in range(len(tickets))])
time = 0
while queue:
    person_id, remaining = queue.popleft()
    time += 1
    if remaining > 1:
        queue.append((person_id, remaining - 1))
    if person_id == k and remaining == 1:
        return time

Why It Happens: The problem description naturally leads to thinking about simulating the actual queue process step-by-step.

The Fix: Recognize that you can calculate the total time mathematically without simulation. The key insight is understanding how many tickets each person can buy before person k finishes.

Pitfall 3: Incorrect Boundary Check

The Mistake: Using strict inequality instead of inclusive comparison for position k:

# INCORRECT - wrong boundary
if i < k:  # Should be i <= k
    total_time += min(ticket_count, tickets[k])

Why It Happens: Confusion about whether person k themselves should be included in the "before or at" category.

The Fix: Remember that person k needs to buy all tickets[k] tickets, so they should be in the first category with i <= k.

Pitfall 4: Edge Case with tickets[k] = 1

The Mistake: Not handling the case where tickets[k] - 1 = 0 for people after position k:

# Potential issue if not careful
if i > k:
    total_time += min(ticket_count, tickets[k] - 1)  # What if tickets[k] = 1?

Why It Happens: When tickets[k] = 1, people after position k get 0 chances (since tickets[k] - 1 = 0).

The Fix: The min() function naturally handles this case correctly. When tickets[k] - 1 = 0, min(ticket_count, 0) = 0, which correctly adds 0 to the total time. No special handling needed, but be aware of this edge case when testing.