Number of Students Unable to Eat Lunch

class Solution {
public:
    int calculateRemainingStudents(vector<int>& students, vector<int>& sandwiches) {
        int zeroPrefCount = 0;
        int onePrefCount = 0;
    
        for (int pref : students) {
            if (pref == 0) zeroPrefCount++;
            else onePrefCount++;
        }
    
        for (int sandwichType : sandwiches) {
            if (sandwichType == 0 && zeroPrefCount == 0) {
                return onePrefCount;
            }
            if (sandwichType == 1 && onePrefCount == 0) {
                return zeroPrefCount;
            }
            if (sandwichType == 0) zeroPrefCount--;
            else onePrefCount--;
        }
        return 0;
    }
};

This article describes a solution in C++ to determine the number of students who are unable to eat lunch due to sandwich preferences mismatches. The solution involves counting students' preferences for two types of sandwiches and matching them against available sandwiches. If a sandwich type cannot be matched with a student who prefers it, because all such students are already served, then all students preferring the other sandwich type are counted as unable to eat.

• Begin by initializing two counters, zeroPrefCount and onePrefCount, to count the number of students who prefer sandwich type 0 and type 1 respectively.
• Iterate through the students vector to populate these preference counts based on the values (0 or 1) encountered.
• Next, loop through the sandwiches vector, which represents the queue of available sandwiches.
  • Check the current sandwich type at the front of the queue. If it is 0 and no students preferring 0 remain (zeroPrefCount equals 0), count all remaining students who prefer type 1 sandwiches (onePrefCount).
  • Similarly, if the sandwich type is 1 and no students preferring 1 remain (onePrefCount equals 0), count all remaining students who prefer type 0 sandwiches (zeroPrefCount).
  • If matches are found (students for the given sandwich type are still present), decrement the corresponding preference count.
• The loop continues until all sandwiches are processed or it becomes impossible to serve any more students due to preference mismatches.
• Finally, return the count of students who were unable to get their preferred sandwich.

This method efficiently calculates the result by tracking preferences and directly matching them with available sandwiches, ending early if a complete mismatch is detected. This guarantees that any student who can be served is accounted for before concluding those who cannot.

class Solution {
    public int distributeSandwiches(int[] students, int[] sandwiches) {
        int countCircle = 0;
        int countSquare = 0;
    
        // Tally preferences of students for each type of sandwich
        for (int preference : students) {
            if (preference == 0) {
                countCircle++;
            } else {
                countSquare++;
            }
        }
    
        // Process each sandwich in the stack
        for (int sandwich : sandwiches) {
            // If no student wants the current circle sandwich
            if (sandwich == 0 && countCircle == 0) {
                return countSquare;
            }
            // If no student wants the current square sandwich
            if (sandwich == 1 && countSquare == 0) {
                return countCircle;
            }
            // Serve the sandwich and decrement the respective count
            if (sandwich == 0) {
                countCircle--;
            } else {
                countSquare--;
            }
        }
        // All students have been served
        return 0;
    }
}

In Java, the provided solution focuses on determining the number of students who cannot eat due to sandwich type mismatches. The code involves managing arrays and effectively serving students with their preferred type of sandwiches until there are no more matches. Here's a breakdown of how the solution operates:

• Initialize counters for each sandwich type: circle (countCircle) and square (countSquare).

• Traverse the array students to tally the preferences:

  • Increment the countCircle for students preferring circular sandwiches.
  • Increment the countSquare for those preferring square sandwiches.

• Iterate through each sandwich in the sandwiches array to serve them in sequence:

  • If the top sandwich on the stack is circular (0) but no students prefer circular sandwiches (countCircle is 0), return the number of remaining students who prefer square sandwiches (countSquare).
  • If the top sandwich is square (1) and no students prefer square (countSquare is 0), return the number of students preferring circular sandwiches (countCircle).
  • Decrement the respective sandwich preference counter upon serving a student.

• If all sandwiches match the student preferences and are served appropriately, return 0, indicating all students have been served.

This algorithm efficiently handles the distribution based on direct comparison and immediate results, ensuring that it can quickly determine when a mismatch occurs and how many students remain unable to be served.

class Solution:
    def studentCount(self, students: List[int], lunch: List[int]) -> int:
        count_circle = 0
        count_square = 0
    
        # Tally preferences for each type
        for student_preference in students:
            if student_preference == 0:
                count_circle += 1
            else:
                count_square += 1
    
        # Distribute sandwiches accordingly
        for sandwich_type in lunch:
    
            # Check availability of circle type sandwiches
            if sandwich_type == 0 and count_circle == 0:
                return count_square
    
            # Check availability of square type sandwiches
            if sandwich_type == 1 and count_square == 0:
                return count_circle
    
            # Update the counts as sandwiches are taken
            if sandwich_type == 0:
                count_circle -= 1
            else:
                count_square -= 1
    
        # All students have received their preferred sandwich
        return 0

The provided Python solution aims to solve the problem where students' preferences for sandwiches are given along with the available types of sandwiches. The goal is to determine how many students will be unable to receive their preferred type of sandwich. Here is a breakdown of how the solution approaches this problem:

• Two counters, count_circle and count_square, are initiated to keep track of the number of students who prefer circle and square sandwiches, respectively.
• Initially, the solution iterates over each student's preference. It increments the corresponding counter based on whether a student prefers a circle or square sandwich.
• Following this, the solution then iterates over each sandwich in the provided order:
  • If it encounters a circle sandwich (sandwich_type == 0) and no more students prefer the circle type (count_circle == 0), the function returns the number of students who still want square sandwiches (count_square).
  • Similarly, if a square sandwich is encountered (sandwich_type == 1) with no remaining preference for square sandwiches among students (count_square == 0), it returns the number of students left wanting circle sandwiches (count_circle).
  • If a preferred sandwich type is available, the corresponding counter is decremented.
• If all sandwiches are distributed without any leftovers in preferences, the function finally returns 0, indicating all students received their preferred types.

This approach effectively assesses the inability of students to get their preferred sandwich based on availability and order, returning the count of those who can't be accommodated.