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

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Number of Students Unable to Eat Lunch?

This problem uses Array, Stack, Queue, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Number of Students Unable to Eat Lunch?

Insight 1: Using a queue (or simulating one with array manipulation) to manage student order is crucial to correctly model the 'go to the end' behavior. Insight 2: The problem can be efficiently solved using a stack (simulated with array manipulation) to represent the sandwich stack and a queue to represent the students. Tracking how many times students rotate helps determine when no student can eat.

Which companies ask LeetCode 1700. Number of Students Unable to Eat Lunch in interviews?

This problem is frequently asked at Flipkart.

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Number of Students Unable to Eat Lunch - LeetCode 1700 Solution

Number of Students Unable to Eat Lunch - Complete Solution Guide

Number of Students Unable to Eat Lunch is LeetCode problem 1700, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

The school cafeteria offers circular and square sandwiches at lunch break, referred to by numbers 0 and 1 respectively. All students stand in a queue. Each student either prefers square or circular sandwiches. The number of sandwiches in the cafeteria is equal to the number of students. The sandwiches are placed in a stack . At each step: If the student at the front of the queue prefers the sandwich on the top of the stack, they will take it and leave the queue. Otherwise, they will leave it and

Detailed Explanation

The problem simulates a school cafeteria lunch line. Students have preferences for either circular (0) or square (1) sandwiches. Sandwiches are in a stack, and students are in a queue. A student at the front of the queue checks the top sandwich. If they like it, they take it and leave. Otherwise, they go to the end of the queue. The process continues until a student cannot get their preferred sandwich; at that point, all remaining students remain hungry. The goal is to count how many students are left unable to eat.

Solution Approach

The provided solutions use a queue to track student preferences and a stack (implicitly handled through array operations in most implementations) for sandwiches. The algorithm iterates, checking if the top sandwich matches the preference of the student at the front of the queue. If a match occurs, both the student and the sandwich are removed; otherwise, the student is moved to the end of the queue. The loop terminates when no students can eat their preferred sandwich. The number of remaining students is the output.

Step-by-Step Algorithm

Step 1: Initialize a counter `unableToEat` to 0.
Step 2: Iterate while the `unableToEat` is less than the total number of students AND there are still sandwiches left.
Step 3: If the top sandwich matches the current student's preference, remove both the student and sandwich, reset `unableToEat`, and proceed.
Step 4: If there is no match, move the student to the end of the queue, increment `unableToEat`, and proceed.
Step 5: Once the loop terminates, calculate the number of remaining students (who cannot eat) and return this value.

Key Insights

Insight 1: Using a queue (or simulating one with array manipulation) to manage student order is crucial to correctly model the 'go to the end' behavior.
Insight 2: The problem can be efficiently solved using a stack (simulated with array manipulation) to represent the sandwich stack and a queue to represent the students. Tracking how many times students rotate helps determine when no student can eat.
Insight 3: The loop terminates when either all sandwiches are eaten or when the number of times a student rotates through the queue equals the total number of students; at this point, no more students can be served.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Stack, Queue, Simulation.

Companies

Asked at: Flipkart.

Number of Students Unable to Eat Lunch | LeetCode 1700

Easy

Array Stack Queue Simulation

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public int countStudents(int[] students, int[] sandwiches) {
        int unableToEat = 0;
        int studentIndex = 0;
        int sandwichIndex = 0;
        int n = students.length;

        while (unableToEat < n && sandwichIndex < n) {
            if (students[studentIndex % n] == sandwiches[sandwichIndex]) {
                unableToEat = 0;
                sandwichIndex++;
                students[studentIndex % n] = -1; 
            } else {
                unableToEat++;
            }
            studentIndex++;
        }

        int count = 0;
        for (int student : students) {
            if (student != -1) {
                count++;
            }
        }

        return count - (n - sandwichIndex);
    }
}

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Problem Info

Description

Problem 1700: Number of Students Unable to Eat Lunch - Detailed explanation and solution approaches.

Topics

Array Stack Queue Simulation

Companies

Flipkart

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1700

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize a counter `unableToEat` to 0.
2
Step 2: Iterate while the `unableToEat` is less than the total number of students AND there are still sandwiches left.
3
Step 3: If the top sandwich matches the current student's preference, remove both the student and sandwich, reset `unableToEat`, and proceed.
4
Step 4: If there is no match, move the student to the end of the queue, increment `unableToEat`, and proceed.
5
Step 5: Once the loop terminates, calculate the number of remaining students (who cannot eat) and return this value.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

In the worst-case scenario, a student might cycle through the queue n times before finding their preferred sandwich. Each cycle involves shifting elements in the array; shifting can take O(n) time. Hence O(n^2). Space complexity is constant because the algorithms use only a few integer variables to keep track of indices and counters. It doesn't scale with the size of input.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a deque

Instead of using an array and manipulating indices to simulate a queue, one could use a `deque` data structure which provides efficient `append` and `pop` operations at both ends. This can improve the performance slightly as we avoid the O(n) array manipulation in the worst case.

Time: O(n) on average, but O(n^2) in the worst case (highly unlikely).Space: O(n) due to the use of deque

Trade-offs: Using a deque offers a slight performance advantage in the average case compared to array manipulation, especially for larger inputs. The space overhead is not significant unless the number of students is very large.

✨

Key Insights

1Insight 1: Using a queue (or simulating one with array manipulation) to manage student order is crucial to correctly model the 'go to the end' behavior.
2Insight 2: The problem can be efficiently solved using a stack (simulated with array manipulation) to represent the sandwich stack and a queue to represent the students. Tracking how many times students rotate helps determine when no student can eat.
3Insight 3: The loop terminates when either all sandwiches are eaten or when the number of times a student rotates through the queue equals the total number of students; at this point, no more students can be served.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Queue

FIFO structure - First In, First Out. Like a line at a store!

Edge Cases

Edge case 1: All students get their preferred sandwiches. The algorithm correctly returns 0 in this case.
Edge case 2: No student gets their preferred sandwich. The algorithm correctly returns the number of students.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the queue. Beginners might forget to move students to the end of the queue when their preference doesn't match the top sandwich. This results in inaccurate tracking of students and sandwiches.

✗Common mistake 2: Infinite loops. Without correctly handling the termination condition (no student can eat, or all sandwiches eaten), the algorithm may run indefinitely. Checking if the `unableToEat` count matches the number of students is crucial to prevent this.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. Imagine assigning tasks to workers where each worker has a preference for certain types of tasks and tasks are available in a limited supply. This algorithm could simulate and optimize the task assignment process.

🏦Real-world scenario 2: Order fulfillment. In a warehouse, orders might be fulfilled using a first-come-first-served approach. However, if certain items are unavailable, the order might be moved to the end of the queue and processed later. This algorithm models such a situation and helps determine the number of unfulfilled orders.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Number of Students Unable to Eat Lunch"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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