What is the time complexity of Maximum Students Taking Exam?

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

What is the space complexity of Maximum Students Taking Exam?

The space complexity is O(2^n).

What data structures are used to solve Maximum Students Taking Exam?

This problem uses Array, Dynamic Programming, Bit Manipulation, Matrix, Bitmask. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Students Taking Exam?

Insight 1: Representing row configurations using bitmasks. Each bit in the mask corresponds to a seat in a row, where a '1' indicates a student is sitting there and a '0' indicates an empty seat. Insight 2: Dynamic Programming is crucial. We can build up the solution row by row, considering the maximum number of students we can seat in each row given the configuration of the previous row.

Which companies ask LeetCode 1349. Maximum Students Taking Exam in interviews?

This problem is frequently asked at SAP.

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Maximum Students Taking Exam - LeetCode 1349 Solution

Maximum Students Taking Exam - Complete Solution Guide

Maximum Students Taking Exam is LeetCode problem 1349, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a m * n matrix seats that represent seats distributions in a classroom. If a seat is broken, it is denoted by '#' character otherwise it is denoted by a '.' character. Students can see the answers of those sitting next to the left, right, upper left and upper right, but he cannot see the answers of the student sitting directly in front or behind him. Return the maximum number of students that can take the exam together without any cheating being possible. Students must be placed in seats i

Detailed Explanation

The problem asks us to find the maximum number of students that can take an exam in a classroom without cheating. The classroom layout is represented by a matrix where '.' denotes an available seat and '#' denotes a broken seat. Students can see the answers of those sitting to their immediate left, right, upper-left, and upper-right. The goal is to maximize the number of students while ensuring no student can see another's answers.

Solution Approach

The solution uses dynamic programming and bit manipulation. It iterates through each row of the seating arrangement. For each row, it considers all possible valid configurations of students (represented as bitmasks). A configuration is valid if no two students are sitting next to each other horizontally and no student is sitting on a broken seat. The algorithm checks for conflicts between the current row's configuration and the previous row's configuration (upper-left and upper-right conflicts). The dynamic programming table `dp` stores the maximum number of students that can be seated up to the current row, given a specific configuration in the previous row. The algorithm builds the dp table row by row and returns the maximum value in the last row of the dp table.

Step-by-Step Algorithm

Step 1: Initialize `broken_masks`: Create an array `broken_masks` of size `m`, where `m` is the number of rows. For each row, create a bitmask where a set bit indicates a broken seat.
Step 2: Initialize `dp`: Create a dynamic programming table `dp` of size `2^n`, where `n` is the number of columns. `dp[mask]` will store the maximum number of students that can be seated up to the previous row, given the configuration `mask` in the previous row. Initialize it to 0 as it represents a 'virtual' row prior to row 0.
Step 3: Iterate through rows: Iterate through each row `r` from 0 to `m - 1`.
Step 4: Create new_dp array: Create a temporary array `new_dp` of size `2^n` to store results for the current row.
Step 5: Iterate through possible masks: Iterate through all possible bitmasks `current_mask` from 0 to `2^n - 1` representing all possible student seating configurations for the current row.
Step 6: Check for validity of current_mask: Check if the `current_mask` is valid. It should not have any students on broken seats and no two students can be adjacent horizontally. Skip to the next mask if its invalid.
Step 7: Calculate num_students_current: If the `current_mask` is valid, count the number of students sitting in this row using bit manipulation (count the number of set bits in the mask).
Step 8: Find best previous state: Iterate through all possible bitmasks `prev_mask` from 0 to `2^n - 1` representing possible student configurations in the previous row.
Step 9: Check compatibility: Check if the `prev_mask` is compatible with the `current_mask`. This means there should be no upper-left or upper-right conflicts between the two masks. Skip to the next prev_mask if they are incompatible.
Step 10: Update max_prev_students: If the `prev_mask` is compatible, update `max_prev_students` with the value in `dp[prev_mask]` if it is larger than the current `max_prev_students`.
Step 11: Update new_dp: After iterating through all compatible `prev_mask` values, store the maximum possible number of students that can be seated up to the current row with this `current_mask` configuration: `new_dp[current_mask] = num_students_current + max_prev_students`.
Step 12: Update dp: After iterating through all possible current_mask values, update the `dp` array with the values in the `new_dp` array: `dp = new_dp`.
Step 13: Find max students: After iterating through all rows, find the maximum value in the `dp` array. This value represents the maximum number of students that can be seated in the classroom without any cheating.

Key Insights

Insight 1: Representing row configurations using bitmasks. Each bit in the mask corresponds to a seat in a row, where a '1' indicates a student is sitting there and a '0' indicates an empty seat.
Insight 2: Dynamic Programming is crucial. We can build up the solution row by row, considering the maximum number of students we can seat in each row given the configuration of the previous row.
Insight 3: Checking for conflicts. A key part of the solution involves efficiently checking for conflicts between students in the current row (horizontal conflicts) and between the current row and the previous row (diagonal conflicts).
Insight 4: A seat being broken is the equivalent of that position not being a candidate for a student to sit in at all. We can precompute which positions those are for each row with another bitmask.

Complexity Analysis

Time Complexity: O(m * n * 2^(2 * n))

Space Complexity: O(2^n)

Topics

This problem involves: Array, Dynamic Programming, Bit Manipulation, Matrix, Bitmask.

Companies

Asked at: SAP.

Maximum Students Taking Exam | LeetCode 1349

Hard

Array Dynamic Programming Bit Manipulation Matrix Bitmask

Time: O(m * n * 2^(2 * n))Space: O(2^n)

Solution Code

Available in:

class Solution {
    public int maxStudents(char[][] seats) {
        int m = seats.length;
        int n = seats[0].length;

        int[] broken_masks = new int[m];
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                if (seats[r][c] == '#') {
                    broken_masks[r] |= (1 << c);
                }
            }
        }

        // dp[mask] represents the max students for a row configuration `mask`
        // in the *previous* row. Initialize for a virtual row -1 before row 0.
        int[] dp = new int[1 << n];

        for (int r = 0; r < m; r++) {
            int[] new_dp = new int[1 << n];
            for (int current_mask = 0; current_mask < (1 << n); current_mask++) {
                // A mask is valid for the current row `r` if:
                // 1. No students are on broken seats.
                if ((current_mask & broken_masks[r]) != 0) {
                    continue;
                }
                // 2. No two students are adjacent horizontally.
                if ((current_mask & (current_mask >> 1)) != 0) {
                    continue;

                // If the current mask is valid, find the best compatible previous state.
                int num_students_current = Integer.bitCount(current_mask);
                int max_prev_students = 0;

                for (int prev_mask = 0; prev_mask < (1 << n); prev_mask++) {
                    // A prev_mask is compatible with current_mask if:
                    // 1. No upper-left conflict: student at (r, c) and (r-1, c-1).
                    if ((current_mask & (prev_mask << 1)) != 0) {
                        continue;
                    }
                    // 2. No upper-right conflict: student at (r, c) and (r-1, c+1).
                    if ((current_mask & (prev_mask >> 1)) != 0) {
                        continue;
                    }

                    max_prev_students = Math.max(max_prev_students, dp[prev_mask]);
                }

                new_dp[current_mask] = num_students_current + max_prev_students;
            }
            dp = new_dp;
        }

        int max = 0;
        for(int val : dp){
            max = Math.max(max, val);
        }
        return max;
    }
}

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Problem Info

Description

Problem 1349: Maximum Students Taking Exam - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Bit Manipulation Matrix Bitmask

Companies

SAP

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1349

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

13 steps

1
Step 1: Initialize `broken_masks`: Create an array `broken_masks` of size `m`, where `m` is the number of rows. For each row, create a bitmask where a set bit indicates a broken seat.
2
Step 2: Initialize `dp`: Create a dynamic programming table `dp` of size `2^n`, where `n` is the number of columns. `dp[mask]` will store the maximum number of students that can be seated up to the previous row, given the configuration `mask` in the previous row. Initialize it to 0 as it represents a 'virtual' row prior to row 0.
3
Step 3: Iterate through rows: Iterate through each row `r` from 0 to `m - 1`.
4
Step 4: Create new_dp array: Create a temporary array `new_dp` of size `2^n` to store results for the current row.
5
Step 5: Iterate through possible masks: Iterate through all possible bitmasks `current_mask` from 0 to `2^n - 1` representing all possible student seating configurations for the current row.
6
Step 6: Check for validity of current_mask: Check if the `current_mask` is valid. It should not have any students on broken seats and no two students can be adjacent horizontally. Skip to the next mask if its invalid.
7
Step 7: Calculate num_students_current: If the `current_mask` is valid, count the number of students sitting in this row using bit manipulation (count the number of set bits in the mask).
8
Step 8: Find best previous state: Iterate through all possible bitmasks `prev_mask` from 0 to `2^n - 1` representing possible student configurations in the previous row.
9
Step 9: Check compatibility: Check if the `prev_mask` is compatible with the `current_mask`. This means there should be no upper-left or upper-right conflicts between the two masks. Skip to the next prev_mask if they are incompatible.
10
Step 10: Update max_prev_students: If the `prev_mask` is compatible, update `max_prev_students` with the value in `dp[prev_mask]` if it is larger than the current `max_prev_students`.
11
Step 11: Update new_dp: After iterating through all compatible `prev_mask` values, store the maximum possible number of students that can be seated up to the current row with this `current_mask` configuration: `new_dp[current_mask] = num_students_current + max_prev_students`.
12
Step 12: Update dp: After iterating through all possible current_mask values, update the `dp` array with the values in the `new_dp` array: `dp = new_dp`.
13
Step 13: Find max students: After iterating through all rows, find the maximum value in the `dp` array. This value represents the maximum number of students that can be seated in the classroom without any cheating.

💡

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(m * n * 2^(2 * n))

💾

Space Complexity

O(2^n)

📝

Why?

The time complexity is derived from iterating through all rows and all possible bitmask combinations for the current and previous row. The space complexity comes from the size of the DP array, which stores the maximum number of students for each possible mask in the previous row.

📖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

Brute Force with Backtracking

Try placing students one by one, exploring all possible combinations and backtracking when a conflict is detected. While simple, this would have very poor performance.

Time: O(2^(m*n)), which is exponential in the number of seatsSpace: O(m*n) due to the recursion depth.

Trade-offs: This approach is highly inefficient and is not practical for any reasonably sized input. It is easier to understand initially, but quickly becomes computationally infeasible.

✨

Key Insights

1Insight 1: Representing row configurations using bitmasks. Each bit in the mask corresponds to a seat in a row, where a '1' indicates a student is sitting there and a '0' indicates an empty seat.
2Insight 2: Dynamic Programming is crucial. We can build up the solution row by row, considering the maximum number of students we can seat in each row given the configuration of the previous row.
3Insight 3: Checking for conflicts. A key part of the solution involves efficiently checking for conflicts between students in the current row (horizontal conflicts) and between the current row and the previous row (diagonal conflicts).
4Insight 4: A seat being broken is the equivalent of that position not being a candidate for a student to sit in at all. We can precompute which positions those are for each row with another bitmask.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty classroom (all '#' seats). The algorithm correctly returns 0 because no students can be seated.
Edge case 2: Single row/column. The algorithm handles cases with only one row or one column correctly.
Edge case 3: All seats are available (all '.' seats). The algorithm correctly maximizes the number of students, respecting the adjacency constraint.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly checking for conflicts. It's crucial to consider both horizontal conflicts in the current row and diagonal conflicts between the current and previous rows. Forgetting either results in incorrect answers.

✗Common mistake 2: Not handling broken seats properly. Students should never be seated on broken seats. This needs to be checked at each row.

✗Common mistake 3: Incorrectly counting the number of set bits within the time-complexity allowed, if that is done in a naive way it could cause a TLE.

✗Common mistake 4: Failing to optimize for the maximum number of students on each row, instead just settling for valid rows.

✅

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: Optimizing seating arrangements in a theater or classroom to maximize capacity while maintaining social distancing or preventing unwanted visibility between attendees.

🏦Real-world scenario 2: Resource allocation in a distributed system where resources cannot be adjacent to each other due to interference or dependencies. This could be used in scheduling tasks on processors, where placing tasks on adjacent processors might cause overheating.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

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

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 29, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Students Taking Exam"?

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

🤖 Need help with this problem?