Solution Explanation: Number of Students Unable to Eat Lunch

This problem involves simulating a queue of students and a stack of sandwiches. Students have preferences (0 or 1 representing sandwich type), and they take sandwiches from the stack if they match their preference; otherwise, they go to the back of the queue. The goal is to find how many students are left unable to eat.

The naive approach of simulating the queue and stack directly is possible but inefficient (O(n^2) in the worst case). A much more efficient solution leverages the fact that once a student doesn't get their preferred sandwich, no further sandwiches can be consumed by any student. This is because the process only changes the order of students in the queue, not the sandwich stack.

Optimized Solution: Counting Preferences

The optimal solution uses counting to track the number of students with each sandwich preference.

1. Initialization:

   • We create a counter (e.g., an array cnt of size 2) to store the counts of students who prefer sandwich type 0 and sandwich type 1.
   • We initialize the counter by iterating through the students array and incrementing the appropriate count for each student's preference.

2. Iterating Through Sandwiches:

   • We iterate through the sandwiches array (representing the stack).
   • For each sandwich type v:
     • If cnt[v] is 0 (no student wants this type), it means we have reached a point where no more students can eat. The number of students who remain unable to eat is the number of students who prefer the other sandwich type (cnt[v ^ 1]). We return this value.
     • Otherwise, we decrement cnt[v] because a student has taken this sandwich.

3. All Students Eat:

   • If the loop completes without finding a case where cnt[v] is 0, it means all students were able to eat. We return 0.

Time and Space Complexity

• Time Complexity: O(n), where n is the number of students (or sandwiches, since they have equal length). We iterate through the sandwiches and students arrays once.
• Space Complexity: O(1). We use a constant-size array (cnt) to store the counts of student preferences. The space used is independent of the input size.

Code Examples (Multiple Languages)

The code examples below implement the optimized counting solution:

Python:

from collections import Counter
 
class Solution:
    def countStudents(self, students: list[int], sandwiches: list[int]) -> int:
        cnt = Counter(students)  #Efficient counter
        for sandwich in sandwiches:
            if cnt[sandwich] == 0:
                return cnt[1 - sandwich] #Other sandwich type
            cnt[sandwich] -= 1
        return 0

Java:

class Solution {
    public int countStudents(int[] students, int[] sandwiches) {
        int[] cnt = new int[2];
        for (int student : students) {
            cnt[student]++;
        }
        for (int sandwich : sandwiches) {
            if (cnt[sandwich] == 0) {
                return cnt[1 - sandwich];
            }
            cnt[sandwich]--;
        }
        return 0;
    }
}

C++:

class Solution {
public:
    int countStudents(vector<int>& students, vector<int>& sandwiches) {
        vector<int> cnt(2, 0);
        for (int student : students) {
            cnt[student]++;
        }
        for (int sandwich : sandwiches) {
            if (cnt[sandwich] == 0) {
                return cnt[1 - sandwich];
            }
            cnt[sandwich]--;
        }
        return 0;
    }
};

JavaScript:

/**
 * @param {number[]} students
 * @param {number[]} sandwiches
 * @return {number}
 */
var countStudents = function(students, sandwiches) {
    let cnt = [0, 0];
    for (let student of students) {
        cnt[student]++;
    }
    for (let sandwich of sandwiches) {
        if (cnt[sandwich] === 0) {
            return cnt[1 - sandwich];
        }
        cnt[sandwich]--;
    }
    return 0;
};

These code examples are concise and efficient, directly reflecting the optimized counting approach described above. They achieve the optimal time and space complexity.