How do you solve Maximum Students Taking Exam on LeetCode?

Maximum Students Taking Exam (LeetCode #1349) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * 2^n * 2^n) time complexity and O(2^n) space complexity. Key insight: Bit manipulation allows efficient representation of seating arrangements.

What is the brute force solution for Maximum Students Taking Exam?

The brute force approach for Maximum Students Taking Exam is Brute Force, with O(2^n * m) time complexity and O(1) space complexity. The brute force approach involves checking every possible combination of students that can be seated in the classroom. This method is straightforward but inefficient, as it does not leverage any optimizations and can lead to exponential time complexity. The optimal Optimal Solution approach improves this to O(m * 2^n * 2^n).

What algorithm or data structure is used to solve Maximum Students Taking Exam?

Maximum Students Taking Exam uses the following concepts: Array, Dynamic Programming, Bit Manipulation, Matrix, Bitmask. The recommended patterns to study are: Bit Manipulation, Dynamic Programming.

What are the interview tips for Maximum Students Taking Exam?

Clearly explain your thought process while coding. Discuss time and space complexity as you go. Practice bit manipulation problems to get comfortable with masks.

What are common mistakes when solving Maximum Students Taking Exam?

Failing to check visibility correctly between rows. Not considering edge cases with broken seats.

#1349

Maximum Students Taking Exam

Hard

ArrayDynamic ProgrammingBit ManipulationMatrixBitmaskBit ManipulationDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^n * m)	O(m * 2^n * 2^n)
Space	O(1)	O(2^n)

💡

Intuition

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

The optimal solution uses dynamic programming with bit masking to efficiently calculate the maximum number of students that can be seated without cheating. By representing the seating arrangement as a bitmask, we can quickly check for valid placements based on previous rows.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP array where dp[i] represents the maximum number of students that can be seated up to row i.
2Step 2: For each row, iterate through all possible bitmasks representing the seating arrangement.
3Step 3: For each bitmask, check if it is valid based on the previous row's bitmask and update the DP array accordingly.

solution.py13 lines

1# Full working Python code
2def maxStudents(seats):
3    m, n = len(seats), len(seats[0])
4    dp = [0] * (1 << n)
5    for i in range(m):
6        new_dp = dp[:]
7        for mask in range(1 << n):
8            if all(seats[i][j] == '.' for j in range(n) if (mask & (1 << j))):
9                for prev in range(1 << n):
10                    if (mask & prev) == 0 and not any((prev & (1 << (j - 1))) and (mask & (1 << j)) for j in range(1, n)):
11                        new_dp[mask] = max(new_dp[mask], dp[prev] + bin(mask).count('1'))
12        dp = new_dp
13    return max(dp)

ℹ

Complexity note: The time complexity is O(m * 2^n * 2^n) because for each row (m), we check all combinations of previous and current row seating arrangements (2^n). The space complexity is O(2^n) due to the DP array storing results for each bitmask.

1Bit manipulation allows efficient representation of seating arrangements.
2Dynamic programming can optimize the selection of valid student placements.

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