How do you solve Number of Students Unable to Eat Lunch on LeetCode?

Number of Students Unable to Eat Lunch (LeetCode #1700) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding student preferences is crucial.

What is the time complexity of Number of Students Unable to Eat Lunch?

The optimal solution for Number of Students Unable to Eat Lunch runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only traverse the students and sandwiches arrays once each, leading to linear time complexity.

What is the brute force solution for Number of Students Unable to Eat Lunch?

The brute force approach for Number of Students Unable to Eat Lunch is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we simulate the process of students trying to take sandwiches. Each student checks the top sandwich and either takes it or moves to the back of the queue. This continues until no student can take the sandwich on top. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Students Unable to Eat Lunch?

Number of Students Unable to Eat Lunch uses the following concepts: Array, Stack, Queue, Simulation. The recommended patterns to study are: Queue, Simulation.

What are the interview tips for Number of Students Unable to Eat Lunch?

Always think about edge cases, like when students have no preferences left. Consider counting or grouping strategies to simplify the problem. Practice simulating processes to understand the flow before optimizing.

What are common mistakes when solving Number of Students Unable to Eat Lunch?

Not considering the case where all sandwiches are taken. Failing to optimize the simulation approach.

#1700

Number of Students Unable to Eat Lunch

Easy

ArrayStackQueueSimulationQueueSimulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of simulating the entire process, we can count how many students prefer each type of sandwich. If the number of students who prefer a type of sandwich exceeds the number of available sandwiches of that type, we can directly calculate how many students will be unable to eat.

⚙️

Algorithm

4 steps

1Step 1: Count the number of students who prefer circular (0) and square (1) sandwiches.
2Step 2: For each sandwich type in the stack, decrement the respective count of students who prefer that type.
3Step 3: If a sandwich type is exhausted, check how many students are left who prefer that type.
4Step 4: The remaining students who cannot eat are those who still prefer a sandwich type that is no longer available.

solution.py9 lines

1def countStudents(students, sandwiches):
2    count = [0, 0]
3    for student in students:
4        count[student] += 1
5    for sandwich in sandwiches:
6        if count[sandwich] == 0:
7            break
8        count[sandwich] -= 1
9    return count[0] + count[1]

ℹ

Complexity note: The time complexity is O(n) because we only traverse the students and sandwiches arrays once each, leading to linear time complexity.

1Understanding student preferences is crucial.
2Simulating the process can be inefficient; counting preferences is more effective.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.