Abstract / Summary

A person needs to buy tickets in a specific order and can only buy one per minute if they have enough money or already own the needed tickets.

Note

The full problem description, including helpful examples, can be found on LeetCode.

This problem is ranked as

and has an acceptance rate of about 73% (400k+ submissions).

Typical patterns to solve this problem:

• Array
• Queue
• Simulation

Every single line of code was written by me, 100% from scratch, between August 2023 and August 2024.
However, two years later I set up this website and used various AI tools to generate most of the walkthrough / explanations below.

See this dedicated page for more details.

Overview

The provided Solution class implements a method timeRequiredToBuy that computes the total number of seconds needed for a person at position k in a ticket queue to finish purchasing all of their tickets. The algorithm iterates once over the tickets list, accumulating the time contributed by each person while respecting the queue dynamics: each second a person at the front buys one ticket and, if they still need more, they move to the end of the line. The implementation leverages the min function to cap contributions and includes a conditional adjustment for people standing behind the target position, effectively handling the “one less” scenario when the target finishes earlier.

Key Steps

Initialization

The method begins by setting a running total to zero and extracting the number of tickets required by the person at index k.

total = 0
need = tickets[k]

These variables establish the baseline for the subsequent accumulation.

Single Pass Accumulation

A for loop iterates over the indices of the tickets list. For each index i, the algorithm determines how many seconds that person will contribute to the overall time. The core calculation uses min(tickets[i], need), which reflects the fact that no one can spend more seconds than the target’s remaining need, because the process stops as soon as the target finishes.

for i in range(len(tickets)):
    buy = min(tickets[i], need)

Adjustment for Positions Behind the Target

When the current index i lies after the target (i > k), the target will have already taken the last ticket before those behind can make their final purchase. Consequently, those behind can contribute at most need - 1 seconds. The code captures this nuance with an inner conditional that re-computes buy using need - 1.

    if i > k:  # one less if behind in queue
        buy = min(tickets[i], need - 1)

Accumulating the Result

Each computed buy value is added to the running total. This step aggregates the time contributed by every person, respecting the earlier conditional adjustment.

    total += buy

Returning the Computed Time

After the loop finishes, the method returns the accumulated total, representing the exact number of seconds required for the person at position k to complete their purchases.

return total

Edge Cases Handled

The algorithm gracefully handles the situation where the target person requires only a single ticket. In that case, need equals 1, and the conditional for indices behind the target reduces their contribution to min(tickets[i], 0), which yields zero, correctly indicating that no further time is added after the target’s purchase.

When the target is at the front of the queue (k == 0), the conditional i > k never triggers, so every person’s contribution is limited by need. This correctly models the scenario where the target finishes before anyone behind them gets another turn.

If the target requires the maximum number of tickets among all participants, the min operation ensures that each person contributes at most their own ticket count, preventing any over-counting beyond the target’s need.

The code also accommodates an empty tickets list implicitly; however, the problem constraints guarantee at least one element, so the method’s logic remains valid for all admissible inputs.

Performance Optimizations

The implementation achieves optimal performance by avoiding nested loops or repeated queue simulations. By collapsing the entire process into a single linear pass, the method eliminates the overhead of repeatedly rotating the queue or decrementing individual ticket counts.

The use of the built-in min function provides a fast, constant-time way to bound each contribution without explicit conditional branching beyond the single if i > k check. This reduces the number of arithmetic operations per iteration.

Storing need once before the loop prevents repeated indexing into the tickets list for the target position, thereby minimizing memory accesses.

Overall, the algorithm runs in linear time relative to the length of the tickets list, with a minimal constant factor due to the straightforward arithmetic and conditional logic.

Imagine a group of friends standing in a long line to buy cinema tickets. Each person wants a certain number of tickets for their family. You are one of the friends in this line, and you know exactly where you are standing. Everyone in the line takes turns buying one ticket at a time. After buying one, they go to the back of the line if they still need more. This keeps happening until you have bought all the tickets you need.

The computer looks at the whole line and counts how many times people step up to the window. To figure this out, the computer looks at how many tickets each person wants. If someone is standing in front of you, they have to finish their turn before you can get yours. If someone is standing behind you, they only get to take a turn if you still need more tickets. The computer adds up every time someone stands at the window until your specific goal is reached. It is like watching a big clock tick every time a person buys a single ticket, counting until you get every ticket you wanted.

Several LLMs ranked my code quality on a scale from 1 (very poor) to 10 (outstanding):

Time: linear
As a rule of thumb, the maximum input size for one second execution time is

• factorial: up to 10 elements
• exponential: up to 20 elements
• polynomial: a few thousand elements
• loglinear: up to 100,000 elements
• linear: a few million elements
• logarithmic: a billion elements
• constant: practically unlimited

I wrote a basic crawler to find other solutions to that problem.
Obviously, I did that after I solved the problem myself.

In general, AlgoMonster and WalkCCC solutions are high-quality and always worth a look.
I haven't checked most of the other solutions, but they might be helpful as well.

And of course, the LeetCode forum is always a good place (or even the best) to look for alternative approaches.

(click on a column header to sort the table)

	author
author	solutions
www	AlgoMonster	C++	Python		Java				TS
	WalkCCC	C++	Python		Java
	leetcode.ca	C++	Python		Java			Go	TS
	Progiez	C++	Python		Java
github	black-shadows	C++
	charles-wangkai				Java
	everthis			JS
	fartem					Ruby
	fsq	C++
	JoshCrozier			JS
	lkwq007		Python
	suxrobGM						C#
	trung-hn		Python
	varunu28				Java
AlgoMonster	C++, Python, Java, TypeScript
WalkCCC	C++, Python, Java
leetcode.ca	C++, Go, Java, Python, TypeScript
Progiez	C++, Java, Python
black-shadows	C++, C++
charles-wangkai	Java
everthis	JavaScript
fartem	Ruby
fsq	C++
JoshCrozier	JavaScript
lkwq007	Python
suxrobGM	C#
trung-hn	Python
varunu28	Java

There are a few extra pages giving you some insights into my setup:

• Tech Stack - my development environment and tools
• AI - how AI helped me create this website (NOT solving the problems)
• Progress - progression over time
• Quality - code quality