Leetcode 1700: Number of Students Unable to Eat Lunch

Input: The input consists of two arrays: 'students' representing the preference of each student (0 for circular, 1 for square), and 'sandwiches' representing the stack of sandwiches (with 0 for circular, 1 for square).

Example: [0, 1, 0, 1], [1, 0, 1, 0]

Constraints:

• 1 <= students.length, sandwiches.length <= 100

• students.length == sandwiches.length

• students[i] and sandwiches[i] are either 0 or 1

Output: The output should be a single integer representing the number of students who are unable to get a sandwich.

Example: 2

Constraints:

• The returned integer should be between 0 and the total number of students.

Goal: To determine how many students are unable to eat based on their preferences and the order of the sandwiches.

Steps:

• Initialize a count for each type of sandwich in the stack and for each student preference.

• Iterate over the queue of students and check if the sandwich at the top of the stack matches the student's preference.

• If the student prefers the sandwich, remove it from the stack and move the student out of the queue.

• If the student does not want the sandwich, move them to the end of the queue.

• Continue the process until no student at the front of the queue can take the sandwich at the top of the stack.

Goal: The array lengths for both students and sandwiches will always be the same, and both will contain only the values 0 and 1.

Steps:

• Both 'students' and 'sandwiches' arrays have a length between 1 and 100.

• The values in 'students' and 'sandwiches' are either 0 or 1.

Assumptions:

• There will always be enough sandwiches for each student at the start, but some may not get one by the end of the process.

• Input: [0, 1, 0, 1], [1, 0, 1, 0]

• Explanation: All students are able to take a sandwich after a few moves, resulting in no students left without food.

• Input: [1, 0, 1, 0, 1, 1], [0, 1, 1, 1, 0, 0]

• Explanation: Two students are unable to get a sandwich by the end of the process due to mismatched preferences.

Approach: The approach to solving this problem involves simulating the sandwich distribution process, keeping track of which students are still waiting for a sandwich.

Observations:

• The process involves moving students around the queue and keeping track of their preferences relative to the sandwiches on top of the stack.

• We can optimize by using a simple queue to manage the students and a counter for the sandwich types to ensure we don't perform unnecessary operations.

Steps:

• Count the number of students preferring each type of sandwich.

• Simulate the process by iterating through the students' queue and checking if the sandwich matches their preference.

• If a student can't get the sandwich they want, move them to the end of the queue.

• Stop when no student at the front of the queue can take the top sandwich.

Empty Inputs:

• The input will never be empty as there is always at least one student and one sandwich.

Large Inputs:

• The solution should efficiently handle arrays up to 100 elements in length.

Special Values:

• If all students prefer the same type of sandwich, the outcome depends on the sandwiches available in the stack.

Constraints:

• The solution must be able to handle up to 100 students and sandwiches efficiently.

int countStudents(vector<int>& A, vector<int>& B) {
    int count[] = {0, 0}, n = A.size(), k;
    for (int a : A)
        count[a]++;
    for (k = 0; k < n && count[B[k]] > 0; ++k)
        count[B[k]]--;
    return n - k;
}

1 : Function Definition

int countStudents(vector<int>& A, vector<int>& B) {

Defines the function `countStudents` that takes two integer vectors, `A` and `B`, representing students and their preferences, and returns the number of students in `B` who cannot find a matching student in `A`.

2 : Variable Initialization

    int count[] = {0, 0}, n = A.size(), k;

Initializes an array `count` to count the occurrences of each student in array `A`, a variable `n` to store the size of array `A`, and a variable `k` for iteration.

3 : Counting Students in A

    for (int a : A)

Iterates through each student in array `A` to count the occurrences of each student using the `count` array.

4 : Counting Students in A

        count[a]++;

For each student `a` in array `A`, increments the corresponding count in the `count` array.

5 : Matching Students

    for (k = 0; k < n && count[B[k]] > 0; ++k)

Iterates through each student in array `B` and checks if the student can be matched with a student in `A` based on the `count` array. Decreases the count if a match is found.

6 : Decreasing Count for Matches

        count[B[k]]--;

If a match is found, decreases the count for student `B[k]` in the `count` array.

7 : Return Unmatched Students

    return n - k;

Returns the number of unmatched students, which is the total number of students (`n`) minus the number of students that were matched (`k`).

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The algorithm processes each student and sandwich once, so the time complexity is O(n), where n is the number of students.

Best Case: O(1)

Worst Case: O(n)

Description: The space complexity is O(n) for storing the queue of students, and O(1) if the queue is not explicitly stored.

Leetcode 1700: Number of Students Unable to Eat Lunch

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