#1700

Easy

auto_awesomeStack-based state management

LeetCode Problem Workspace

Number of Students Unable to Eat Lunch

Simulate a queue of students with sandwich preferences and determine how many can't eat based on available sandwiches.

Problem Statement

You are given an array of students where each student prefers either a circular sandwich (denoted by 0) or a square sandwich (denoted by 1). The cafeteria offers an equal number of sandwiches in a stack, with each sandwich also being either circular (0) or square (1). Students take turns attempting to grab the top sandwich from the stack. If a student’s preference matches the sandwich type, they take it and leave the line. If not, they move to the back of the line. This process repeats until no student is able to take the top sandwich.

Your goal is to determine how many students remain unable to eat. The process stops when no student is willing to take the top sandwich anymore, indicating a failure to satisfy the students’ needs.

Examples

Example 1

Input: students = [1,1,0,0], sandwiches = [0,1,0,1]

Output: 0

Front student leaves the top sandwich and returns to the end of the line making students = [1,0,0,1].
Front student leaves the top sandwich and returns to the end of the line making students = [0,0,1,1].
Front student takes the top sandwich and leaves the line making students = [0,1,1] and sandwiches = [1,0,1].
Front student leaves the top sandwich and returns to the end of the line making students = [1,1,0].
Front student takes the top sandwich and leaves the line making students = [1,0] and sandwiches = [0,1].
Front student leaves the top sandwich and returns to the end of the line making students = [0,1].
Front student takes the top sandwich and leaves the line making students = [1] and sandwiches = [1].
Front student takes the top sandwich and leaves the line making students = [] and sandwiches = []. Hence all students are able to eat.

Example 2

Input: students = [1,1,1,0,0,1], sandwiches = [1,0,0,0,1,1]

Output: 3

Example details omitted.

Constraints

1 <= students.length, sandwiches.length <= 100
students.length == sandwiches.length
sandwiches[i] is 0 or 1.
students[i] is 0 or 1.

Solution Approach

Simulating Sandwich Distribution

We can simulate the process using a queue for the students and a stack for the sandwiches. For each iteration, check if the student at the front of the queue wants the sandwich at the top of the stack. If not, move that student to the back. Repeat until either all students are served or no student can take the top sandwich.

Handling Edge Cases

Edge cases include situations where all students have the same preference, and no one will take the sandwich type at the top. Another edge case is when the number of students with a certain preference exceeds the number of available sandwiches of that type, causing some students to be unable to eat.

Optimal Time Complexity

The algorithm runs efficiently with O(n) time complexity since each student is moved through the queue at most once, and the sandwich stack operations are O(1). The solution ensures that every student is processed in linear time.

Complexity Analysis

Metric	Value
Time	O(n + m)
Space	O(1)

The time complexity is O(n) where n is the number of students (or sandwiches). Since each student can be moved at most once through the queue and sandwiches are served in constant time, the solution is optimal. The space complexity is O(1) as we are only using the input arrays for simulation.

What Interviewers Usually Probe

A candidate should be able to efficiently simulate the problem with a queue and stack.
Look for an understanding of stack-based state management, which is central to solving this problem.
The candidate should avoid unnecessary operations or memory usage when simulating the process.

Common Pitfalls or Variants

Common pitfalls

Over-complicating the simulation with unnecessary data structures or loops.
Not handling the case when all students have the same sandwich preference, causing an infinite loop of students not getting food.
Ignoring the constraint of student preferences and prematurely stopping the simulation without checking if all students can be served.

Follow-up variants

Change the number of sandwich types or preferences, introducing additional sandwich types.
Modify the queue order of students, such as processing them in reverse or with other variations in the way preferences are set.
Introduce scenarios where the number of sandwiches exceeds the number of students, testing how the solution handles surplus sandwiches.

FAQ

What is the main pattern used in the 'Number of Students Unable to Eat Lunch' problem?

The problem relies on simulating a queue of students and a stack of sandwiches, where each student tries to eat a sandwich based on their preference. This stack-based state management pattern helps manage the process efficiently.

How do I handle the scenario where no student can take the top sandwich?

In this case, the simulation should stop when all students have returned to the end of the queue without taking the sandwich, indicating that no one can be served.

How can I optimize the solution for 'Number of Students Unable to Eat Lunch'?

The key is to manage the queue and stack efficiently. Avoid redundant checks and ensure each student is processed only once, resulting in an O(n) solution.

What is the time complexity of the 'Number of Students Unable to Eat Lunch' solution?

The time complexity is O(n), where n is the number of students, as each student is moved at most once through the queue, and sandwich stack operations are O(1).

Can the solution be extended to handle more than two types of sandwiches?

Yes, the solution can be adapted by modifying the queue and stack management to accommodate multiple sandwich types and preferences.

terminal

Solution

Solution 1: Counting

We observe that the positions of the students can be adjusted, but the positions of the sandwiches cannot be adjusted. That is to say, if the sandwich in front is not taken, then all the sandwiches behind cannot be taken.

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class Solution:
    def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
        cnt = Counter(students)
        for v in sandwiches:
            if cnt[v] == 0:
                return cnt[v ^ 1]
            cnt[v] -= 1
        return 0

Continue Practicing

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