LeetCode 1700 - Number of Students Unable to Eat Lunch

This problem simulates a school cafeteria where students line up to take sandwiches from a stack. Each student has a preference for either a circular sandwich (0) or a square sandwich (1).

leetcodeeasyarraystackqueuesimulation

LeetCode Problem 1700

Difficulty: 🟢 Easy
Topics: Array, Stack, Queue, Simulation

Solution

Problem Understanding

This problem simulates a school cafeteria where students line up to take sandwiches from a stack. Each student has a preference for either a circular sandwich (0) or a square sandwich (1). The sandwiches are arranged in a stack where the first element of the sandwiches array represents the top of the stack. Students form a queue, and the first student in the queue checks if they like the sandwich on top of the stack. If they do, they take it and leave the queue. If not, they move to the back of the queue. This continues until the top sandwich cannot be eaten by any student in the queue.

The input consists of two arrays: students and sandwiches. Both arrays have the same length, with values 0 or 1. The output is the number of students who cannot eat any sandwich when the process terminates.

Constraints tell us the input size is small (maximum 100 students), so algorithms with moderate complexity will run efficiently. Edge cases to consider include all students preferring the same sandwich, all sandwiches being the same, or a mixture where some students cannot be satisfied.

Key observations are that once a sandwich type has no students left who prefer it, the process stops immediately. This insight allows us to avoid simulating every queue rotation.

Approaches

Brute Force

The brute-force method directly simulates the process described in the problem. You maintain a queue for students and a stack for sandwiches. At each iteration, check the front student against the top sandwich. If the student does not like it, move them to the back of the queue and continue. Keep a counter of consecutive students who have refused the top sandwich. If this counter equals the length of the queue, no student wants the top sandwich, and the process stops.

This approach is correct because it directly follows the problem statement, but it involves moving students repeatedly in a queue, giving an O(n^2) worst-case complexity. For small input sizes (≤100), this is acceptable, but it is not optimal.

Optimal Approach

The optimal approach relies on counting rather than simulating. Count how many students prefer circular (0) and square (1) sandwiches. Then iterate through the sandwich stack from top to bottom. If a sandwich's type has at least one student preferring it, decrement that count. If no students prefer the current sandwich, stop the process because remaining students cannot eat.

This approach works because the exact order of students in the queue does not matter once we know the counts of preferences. It eliminates the need for repeated queue rotations.

Approach Time Complexity Space Complexity Notes
Brute Force O(n^2) O(n) Simulates the queue and stack exactly as described
Optimal O(n) O(1) Counts student preferences and iterates through sandwiches without queue simulation

Algorithm Walkthrough

  1. Initialize two counters count0 and count1 to count students who prefer circular (0) and square (1) sandwiches.
  2. Iterate through the students array and increment count0 or count1 based on the student’s preference.
  3. Iterate through the sandwiches stack from top to bottom.
  4. For each sandwich, check if there is a student who prefers it (check count0 for 0 and count1 for 1).
  5. If a student prefers the sandwich, decrement the corresponding counter. This represents the student taking the sandwich.
  6. If no student prefers the current sandwich, stop iterating. The remaining sandwiches cannot be taken, and the remaining students are unable to eat.
  7. Return the total remaining students by summing the remaining counts.

Why it works: The algorithm works because it relies on the invariant that only the counts of students matter, not their order. As long as there is at least one student who prefers a sandwich type, the process can continue. The first sandwich that cannot be taken halts the process, which aligns perfectly with the problem statement.

Python Solution

from typing import List

class Solution:
    def countStudents(self, students: List[int], sandwiches: List[int]) -> int:
        count0 = students.count(0)
        count1 = students.count(1)
        
        for s in sandwiches:
            if s == 0:
                if count0 > 0:
                    count0 -= 1
                else:
                    break
            else:
                if count1 > 0:
                    count1 -= 1
                else:
                    break
        
        return count0 + count1

Implementation Explanation: First, we count the number of students preferring each type of sandwich. Then, we iterate over the sandwiches from top to bottom. If a sandwich can be taken by some student, we decrement the count. If no student can take the current sandwich, we stop. The remaining sum of counts gives the number of students unable to eat.

Go Solution

func countStudents(students []int, sandwiches []int) int {
    count0, count1 := 0, 0
    for _, s := range students {
        if s == 0 {
            count0++
        } else {
            count1++
        }
    }

    for _, s := range sandwiches {
        if s == 0 {
            if count0 > 0 {
                count0--
            } else {
                break
            }
        } else {
            if count1 > 0 {
                count1--
            } else {
                break
            }
        }
    }
    return count0 + count1
}

Go-Specific Notes: Go does not have a built-in count method for slices, so we iterate manually. The approach is otherwise identical to Python, using integer counters. There are no nil issues because the slices are guaranteed to have equal non-zero lengths.

Worked Examples

Example 1: students = [1,1,0,0], sandwiches = [0,1,0,1]

Step Sandwich Count0 Count1 Action
1 0 2 2 count0 > 0, take sandwich, count0 = 1
2 1 1 2 count1 > 0, take sandwich, count1 = 1
3 0 1 1 count0 > 0, take sandwich, count0 = 0
4 1 0 1 count1 > 0, take sandwich, count1 = 0

All students have eaten, return 0.

Example 2: students = [1,1,1,0,0,1], sandwiches = [1,0,0,0,1,1]

Step Sandwich Count0 Count1 Action
1 1 2 4 take sandwich, count1 = 3
2 0 2 3 take sandwich, count0 = 1
3 0 1 3 take sandwich, count0 = 0
4 0 0 3 cannot take, stop

Remaining students = count0 + count1 = 0 + 3 = 3.

Complexity Analysis

Measure Complexity Explanation
Time O(n) Count students once and iterate sandwiches once
Space O(1) Only two counters used, independent of input size

The algorithm does not require extra space proportional to input length and avoids repeated queue rotations.

Test Cases

# Provided examples
assert Solution().countStudents([1,1,0,0], [0,1,0,1]) == 0  # all students can eat
assert Solution().countStudents([1,1,1,0,0,1], [1,0,0,0,1,1]) == 3  # three students cannot eat

# Edge cases
assert Solution().countStudents([0,0,0], [1,1,1]) == 3  # no student likes any sandwich
assert Solution().countStudents([1,1,1], [1,1,1]) == 0  # all students like all sandwiches
assert Solution().countStudents([0,1,0,1], [0,1,0,1]) == 0  # alternating preferences and sandwiches
assert Solution().countStudents([0], [0]) == 0  # single student can eat
assert Solution().countStudents([1], [0]) == 1  # single student cannot eat
Test Why
[1,1,0,0], [0,1,0,1] Checks normal simulation where all students eat
[1,1,1,0,0,1], [1,0,0,0,1,1] Checks stopping condition when sandwiches cannot be eaten
[0,0,0], [1,1,1] All students unable to eat
[1,1,1], [1,1,1] All students can eat
[0,1,0,1], [0,1,0,1] Alternating pattern
[0], [0] Minimum input size where student can eat
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