Introduction

In this blog post, we will explore how to solve the LeetCode problem "1349" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 in good condition. Example 1: Input: seats = [["#",".","#","#",".","#"], [".","#","#","#","#","."], ["#",".","#","#",".","#"]] Output: 4 Explanation: Teacher can place 4 students in available seats so they don't cheat on the exam. Example 2: Input: seats = [[".","#"], ["#","#"], ["#","."], ["#","#"], [".","#"]] Output: 3 Explanation: Place all students in available seats. Example 3: Input: seats = [["#",".",".",".","#"], [".","#",".","#","."], [".",".","#",".","."], [".","#",".","#","."], ["#",".",".",".","#"]] Output: 10 Explanation: Place students in available seats in column 1, 3 and 5. Constraints: seats contains only characters '.' and'#'. m == seats.length n == seats[i].length 1 <= m <= 8 1 <= n <= 8

Explanation

Here's the solution to the problem, addressing the constraints and optimizing for efficiency.

• State Representation: The core idea is to use dynamic programming with bitmasking. A state is represented by a bitmask where each bit corresponds to a seat in a row. A '1' indicates a student is seated, and '0' indicates the seat is empty.

• Dynamic Programming: We iterate through each row, and for each possible configuration (bitmask) of students in that row, we check if the configuration is valid (no adjacent students, only placing students in available seats). Then, we calculate the maximum number of students that can be seated up to that row, considering all valid configurations in the previous row such that no students can cheat.

• Optimization: The small constraints on m and n (1 <= m, n <= 8) allow us to efficiently explore all possible states (bitmasks).

• Runtime Complexity: O(m * 2^(2n)), where m is the number of rows and n is the number of columns. This is due to iterating through each row (m) and all possible combinations of students in the current and previous rows (2^n for each row, and checking compatibility involves 2^n states of the previous row.

• Storage Complexity: O(2n), due to the space needed to store results from the previous row to calculate optimal seating arrangements.

Code

    def maxStudents(seats):
    m = len(seats)
    n = len(seats[0])

    dp = {}  # Stores the maximum students up to a row with a certain configuration

    def is_valid(mask, row):
        """Checks if a seating arrangement (mask) is valid for a given row."""
        for i in range(n):
            if (mask >> i) & 1:  # If a student is seated at this position
                if seats[row][i] == '#':
                    return False  # Seat is broken
                if i > 0 and (mask >> (i - 1)) & 1:
                    return False  # Adjacent student to the left

        return True

    def count_students(mask):
        """Counts the number of students in a given mask."""
        count = 0
        for i in range(n):
            if (mask >> i) & 1:
                count += 1
        return count

    for row in range(m):
        new_dp = {}
        for mask in range(1 << n):  # Iterate through all possible configurations of the current row
            if not is_valid(mask, row):
                continue

            num_students = count_students(mask)

            if row == 0:
                new_dp[mask] = num_students
            else:
                for prev_mask in dp:
                    # Check if students in the current row can cheat from the previous row
                    can_cheat = False
                    for i in range(n):
                        if (mask >> i) & 1:
                            # Check upper left
                            if i > 0 and (prev_mask >> (i - 1)) & 1:
                                can_cheat = True
                                break
                            # Check upper right
                            if i < n - 1 and (prev_mask >> (i + 1)) & 1:
                                can_cheat = True
                                break

                    if not can_cheat:
                        new_dp[mask] = max(new_dp.get(mask, 0), dp[prev_mask] + num_students)
        dp = new_dp

    if not dp:
        return 0

    return max(dp.values())