Number of Students Unable to Eat Lunch

#1767 Most AskedEasy

Number of Students Unable to Eat Lunch #1767 Most Asked

Easy

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Arrays

The school cafeteria offers circular and square sandwiches at lunch break, referred to by numbers 0 and 1 respectively. All students stand in a queue. Each student either prefers square or circular sandwiches.

The number of sandwiches in the cafeteria is equal to the number of students. The sandwiches are placed in a stack. At each step:

If the student at the front of the queue prefers the sandwich on the top of the stack, they will take it and leave the queue.
Otherwise, they will leave it and go to the queue's end.

This continues until none of the queue students want to take the top sandwich and are thus unable to eat.

You are given two integer arrays students and sandwiches where sandwiches[i] is the type of the i^th sandwich in the stack (i = 0 is the top of the stack) and students[j] is the preference of the j^th student in the initial queue (j = 0 is the front of the queue). Return the number of students that are unable to eat.

Example 1:

Input: students = [1,1,0,0], sandwiches = [0,1,0,1]
Output: 0 
Explanation:
- 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

Constraints:

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible values for elements in the `students` and `sandwiches` arrays (e.g., are they only 0 and 1, or can they be other integers)?
What is the maximum size of the `students` and `sandwiches` arrays? Can they be empty?
If all students eventually get their preferred sandwich, what should the function return (0, or some other value)?
Are the `students` and `sandwiches` arrays guaranteed to have the same length?
Is the order of preference in the `students` array important, or only the *number* of students with each preference?

Brute Force Solution

Approach

The brute force approach to this problem is like simulating the lunch line process exactly as described, following the given rules, until everyone either gets lunch or is stuck. We essentially play out the scenario step-by-step to see what happens in the end.

Here's how the algorithm would work step-by-step:

Start with the initial line of students and the initial stack of sandwiches.
Take the first student in line. If the student's preference matches the top sandwich, give the sandwich to the student and remove both from their respective lines.
If the student's preference does not match the top sandwich, move the student to the end of the line.
Repeat the above two steps until one of two things happens: either all students have eaten, or no students are eating even after going through the line.
If all students have eaten, the answer is zero.
If students are still in line and none of them can take a sandwich, the answer is the number of students remaining in line.

Code Implementation

def count_students_unable_to_eat(students, sandwiches):
    student_queue = students[:] # Create a copy to avoid modifying the original
    sandwich_stack = sandwiches[:]

    while student_queue:
        student_preference = student_queue[0]
        top_sandwich = sandwich_stack[0]

        # Check if the student can eat the top sandwich
        if student_preference == top_sandwich:
            student_queue.pop(0)
            sandwich_stack.pop(0)

            # If no sandwiches are left, all students have eaten
            if not sandwich_stack:
                return 0
        else:
            # Move the student to the end of the queue if they can't eat
            student_queue.append(student_queue.pop(0))

            # Check if no student can eat the sandwiches
            if all(student_queue[i] != sandwich_stack[0] for i in range(len(student_queue))):
                # If no one can eat, break the loop
                break

    # Return the number of students unable to eat
    return len(student_queue)

Big(O) Analysis

Time Complexity

O(n²) – The while loop continues until either the student or sandwich list is empty or a full rotation of the student list happens without anyone eating. In the worst-case scenario, each student in the students array (of size n) might need to traverse the entire line before being unable to eat, which requires checking up to n sandwiches in the sandwich array. Thus, each of the n students may require going through n sandwiches resulting in approximately n * n operations. Hence, the Big O time complexity is O(n²).

Space Complexity

O(1) – The brute force approach primarily modifies the input lists (students and sandwiches) in place or reassigns them. No auxiliary data structures, such as temporary lists or hash maps, are created that scale with the input size N (where N is the number of students). The algorithm uses only a few constant-size variables for loop counters and comparisons, irrespective of N. Therefore, the space complexity is constant.

Optimal Solution

Approach

The core idea is to simulate the lunch line process efficiently. Instead of repeatedly searching, we keep track of how many students like each type of sandwich and how many of each sandwich type are available.

Here's how the algorithm would work step-by-step:

Count how many students prefer each type of sandwich (type 0 and type 1).
Count how many sandwiches of each type are available (type 0 and type 1).
Go through the line of students one by one.
If the student at the front of the line wants the type of sandwich that's currently available, then decrease the count for both the students and the sandwiches of that type.
If the student at the front of the line wants a type of sandwich that isn't available, then put them at the back of the line, but before doing so, check if the loop will keep going forever, that is, no sandwich can satisfy students.
If the loop will go on forever and there is no satisfaction to the needs of any student, then count how many students are left in the line, which is the answer.

Code Implementation

def count_students(students, sandwiches):    student_prefers_type_zero = students.count(0)
    student_prefers_type_one = students.count(1)
    sandwich_type_zero = sandwiches.count(0)
    sandwich_type_one = sandwiches.count(1)
    
    student_queue = students[:] # Create copy

    while student_queue:
        current_student_preference = student_queue[0]

        # Check if student can eat current sandwich
        if (current_student_preference == 0 and sandwich_type_zero > 0) or \
           (current_student_preference == 1 and sandwich_type_one > 0):
            if current_student_preference == 0:
                sandwich_type_zero -= 1
                student_prefers_type_zero -= 1
            else:
                sandwich_type_one -= 1
                student_prefers_type_one -= 1
            student_queue.pop(0)
        else:
            # Check if no student can eat
            if (student_prefers_type_zero > 0 and sandwich_type_zero == 0) or \
               (student_prefers_type_one > 0 and sandwich_type_one == 0):
                # No student will ever be able to eat
                return len(student_queue)
            
            student_queue.append(student_queue.pop(0))

    return 0

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the list of students (n) multiple times. In the worst-case scenario, where no student can find a matching sandwich, each student might be moved to the back of the line repeatedly. Before moving a student, a check is performed to determine if a match is possible in the future. This check examines sandwich counts versus student preferences. The outer loop continues as long as there are students and sandwiches available. Thus, the number of iterations is potentially proportional to n * n, giving a time complexity of O(n²).

Space Complexity

O(1) – The algorithm utilizes a fixed number of integer variables to count the number of students and sandwiches of each type (0 and 1). These counts occupy constant space, irrespective of the number of students (N). No auxiliary data structures that scale with the input size are created. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Empty students or sandwiches array	Return 0 since no students can eat lunch.
Students array is significantly larger than sandwiches array	All sandwiches will be consumed but some students will be left out; return the length of the students array.
Sandwiches array is significantly larger than the students array	All students will eat and some sandwiches will be left, return 0.
All students prefer the same sandwich, but there's only one of that type available	The solution correctly handles this as the rest of the students will be unable to eat.
Infinite loop scenario: students keep cycling without matching sandwiches	Track the number of consecutive failed attempts; if it equals the number of students, exit the loop.
Large input arrays (performance implications)	Using a queue and comparing the front element avoids unnecessary iterations and provides O(n) time complexity.
Sandwiches and students arrays of different lengths	The algorithm works correctly even with different array sizes by iterating either until students all eat or sandwiches all are taken
Students array is all 0s and sandwiches array starts with 1	The algorithm should correctly identify that all students are unable to eat.

Empty students or sandwiches array

How to Handle:

Return 0 since no students can eat lunch.

Students array is significantly larger than sandwiches array

How to Handle:

All sandwiches will be consumed but some students will be left out; return the length of the students array.

Sandwiches array is significantly larger than the students array

How to Handle:

All students will eat and some sandwiches will be left, return 0.

All students prefer the same sandwich, but there's only one of that type available

How to Handle:

The solution correctly handles this as the rest of the students will be unable to eat.

Infinite loop scenario: students keep cycling without matching sandwiches

How to Handle:

Track the number of consecutive failed attempts; if it equals the number of students, exit the loop.

Large input arrays (performance implications)

How to Handle:

Using a queue and comparing the front element avoids unnecessary iterations and provides O(n) time complexity.

Sandwiches and students arrays of different lengths

How to Handle:

The algorithm works correctly even with different array sizes by iterating either until students all eat or sandwiches all are taken

Students array is all 0s and sandwiches array starts with 1

How to Handle:

The algorithm should correctly identify that all students are unable to eat.