Date: 2026-06-06
Time: 19:30
This file solves LeetCode 2073: Time Needed to Buy Tickets. It models a queue where each person buys one ticket per second, then goes to the back of the line. The function computes how many seconds elapse before the person at index k finishes buying all their tickets.
timetobuy_tickets(tickets, k) -> intThe single exported function. Instead of simulating the queue round by round (O(n * max(tickets))), it computes the answer in a single pass (O(n)) by reasoning about how many times each person gets to buy before person k finishes.
Contract: Given a list of positive integers tickets and a valid index k, returns the total seconds until tickets[k] reaches zero.
Closed-form simulation avoidance. The naive approach is to loop through the queue repeatedly, decrementing each person's count. This solution replaces that simulation with a mathematical observation:
i standing before or at k (i <= k) gets to buy in every round that person k does, including k's final round. So they contribute min(tickets[i], tickets[k]) seconds.i standing after k (i > k) misses k's final round — by the time it's their turn in that last round, k has already finished. So they contribute min(tickets[i], tickets[k] - 1) seconds.This is a common LeetCode idiom: replacing an explicit simulation with per-element contribution analysis.
Imports: None — pure function with no external dependencies.
Imported by: The "Imported By" list in the prompt is misleading — those are test files across the entire repo that likely share a common test runner or import pattern, not files that actually call timetobuytickets. The real consumer is time-needed-to-buy-tickets/testsolution.py.
1. Initialize total = 0.
2. Iterate over each person (i, t) in tickets.
3. For i <= k: add min(t, tickets[k]) — this person buys in all rounds up to and including k's last.
4. For i > k: add min(t, tickets[k] - 1) — this person doesn't get a turn in k's final round.
5. Return total.
k must be a valid index into tickets. No bounds checking is performed.tickets[k] >= 1. If tickets[k] == 0, the tickets[k] - 1 term becomes -1, which would produce incorrect results (negative contributions from the min).<= k get one more round than people at indices > k.None. The function trusts its inputs per LeetCode conventions. Invalid inputs (empty list, out-of-bounds k, non-positive ticket counts) produce undefined behavior rather than exceptions.
time-needed-to-buy-tickets/test_solution.py — Validate the edge cases: k at boundaries, single-element list, all-equal ticketstime-needed-to-buy-tickets/plan.md — See the initial problem decomposition and whether simulation was considered before the O(n) approachtime-needed-to-buy-tickets/review.md — Check if the review flagged the zero-ticket edge case or discussed the simulation-vs-closed-form tradeoffsimulation-to-formula-pattern — Many queue/circular problems (e.g., number-of-students-unable-to-eat-lunch, pass-the-pillow) use the same pattern of replacing simulation with per-element contributionpass-the-pillow/solution.py:passThePillow — Another circular-traversal problem that likely uses modular arithmetic instead of simulationtickets-o-n-single-pass — The solution runs in O(n) time with a single pass over the array, avoiding O(n * max(tickets)) simulationsplit-at-k-boundary — People at indices <= k contribute min(t, tickets[k]) and people after k contribute min(t, tickets[k] - 1), reflecting that post-k people miss the final roundtickets-k-ge-1-assumed — The function assumes tickets[k] >= 1; a zero value would cause tickets[k] - 1 = -1 and incorrect results from the min for post-k elementsno-imports-pure-function — The solution has zero dependencies and is a pure function of its inputs