algoadvance

You are given two integer arrays students and sandwiches where:

• students[i] is the type of the i-th student’s preferred sandwich (0 – the student prefers square sandwiches, or 1 – the student prefers circular sandwiches).
• sandwiches[j] is the type of the j-th sandwich in the stack (i.e., the j-th sandwich is the one on the top of the stack).

The students are standing in a queue. Each student can either take the sandwich on the top of the stack or refuse it and go to the end of the queue. This continues until none of the students want to take the top sandwich and are thus unable to eat lunch. You need to return the number of students that are unable to eat lunch.

Clarifying Questions

1. What are the constraints on the lengths of the students and sandwiches arrays?
   • This is essential for understanding time complexity considerations.
2. Can there be a student type not equal to 0 or 1, or a sandwich type not equal to 0 or 1?
   • No. According to the problem description, only 0 and 1 are valid types.
3. Is the number of students always equal to the number of sandwiches?
   • Yes. This is implied by the problem description.

Strategy

To solve this problem, we’ll follow these steps:

1. Track the count of students preferring each type of sandwich (0 and 1).
2. Iterate through the sandwiches stack, and for each sandwich, check if there are students available who prefer that type.
3. If a student prefers the sandwich on top, serve it and update the count of students for that type.
4. If at any point a sandwich cannot be served because there are no students who prefer it, count the remaining students and return that count.

Code

def countStudents(students, sandwiches):
    # Count the number of students who prefer each type of sandwich
    student_count = [students.count(0), students.count(1)]

    for sandwich in sandwiches:
        # If there is no student preferring the current type of sandwich, break the loop.
        if student_count[sandwich] == 0:
            break
        # Otherwise, serve the sandwich and reduce the count of the corresponding student type.
        else:
            student_count[sandwich] -= 1

    # The remaining students are the ones who can't get their preferred sandwich
    return sum(student_count)

# Example usage:
students = [1, 1, 0, 0]
sandwiches = [0, 1, 0, 1]
print(countStudents(students, sandwiches))  # Output: 0

Time Complexity

• Initialization: Counting the number of 0s and 1s in the students list takes O(n) time, where n is the number of students.
• Iteration: Iterating through the sandwiches stack and updating the counts takes O(n) time.

Thus, the overall time complexity is O(n).

This solution efficiently handles the problem within the constraints.

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