Date: 2026-06-06
Time: 18:23
This file solves LeetCode 1700: Number of Students Unable to Eat Lunch. The problem simulates a cafeteria queue: students line up for sandwiches, each student prefers either circular (0) or square (1) sandwiches, and the sandwich stack is served top-first. A student who doesn't like the top sandwich goes to the back of the line. The process deadlocks when no remaining student wants the top sandwich.
countStudents(students, sandwiches) -> intThe main solver. Takes two lists of 0s and 1s and returns how many students can't eat.
Contract: students contains preferences (0 or 1), sandwiches is an ordered stack (index 0 = top). Returns a non-negative integer in [0, len(students)].
mintimetoremoveballoonsA module-level alias pointing to countStudents. Likely exists because the test harness or a related problem variant expects this name.
Counter-based simulation avoidance. The naive approach simulates the queue rotation — O(n^2) in the worst case. This solution recognizes that the queue order doesn't matter: what matters is *whether any student remaining wants the current top sandwich*. A Counter tracks how many students want each type. This reduces the problem to a single linear scan of the sandwich stack.
This is a common idiom in this repo's solutions: replace simulation with counting/frequency analysis when the order of consumption is determined by supply, not queue position.
Imports: collections.Counter — the only dependency. No custom modules.
Imported by: The test_solution.py in this same directory. The massive "Imported By" list in the metadata is an artifact of the repo's test infrastructure — those test files likely share a common import pattern, not a direct dependency on this module.
1. Count how many students want type-0 vs type-1 sandwiches.
2. Iterate through sandwiches top-to-bottom.
3. For each sandwich s: if count[s] == 0, no remaining student wants it — the queue is deadlocked. Return the sum of all remaining students (count[0] + count[1]).
4. Otherwise, decrement count[s] (one student of that preference eats) and continue.
5. If all sandwiches are consumed, return 0.
count[0] + count[1] equals the number of students still waiting.None. The function assumes valid input per LeetCode constraints (non-empty lists of 0s and 1s, equal lengths). No bounds checking, type validation, or exception handling — appropriate for a competitive programming solution.
number-of-students-unable-to-eat-lunch/test_solution.py — See which edge cases are tested (empty inputs, all-same preferences, immediate deadlock)number-of-students-unable-to-eat-lunch/plan.md — The planning doc may explain why simulation was rejected in favor of countingtime-needed-to-buy-tickets/solution.py:timeNeedsToBuyTickets — Another queue-simulation problem likely solved with a similar "skip the simulation" insightcounter-over-simulation — Survey other solutions in this repo that replace queue/stack simulation with frequency counting (e.g., lemonade-change, implement-queue-using-stacks)counter-deadlock-detection — The algorithm detects deadlock by checking count[s] == 0, which is equivalent to "no student in the queue wants the current top sandwich," regardless of queue ordering.linear-time-guarantee — The solution runs in O(n) time and O(1) space (the Counter has at most 2 keys), compared to O(n^2) for naive simulation.queue-order-irrelevance — The correctness relies on the insight that queue position determines eating order but not eating possibility — any student of the matching type will eventually rotate to the front.alias-is-identity — mintimetoremoveballoons is the same function object as countStudents, not a wrapper — calling either has identical behavior.