LeetCode 3450: Maximum Students on a Single Bench Solution

Master LeetCode problem 3450 (Maximum Students on a Single Bench), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3450. Maximum Students on a Single Bench

Easy

Array Hash Table

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming with bitmasking. We will iterate over each row of students and compute the maximum number of students that can sit on a bench without violating the given conditions. We will use a bitmask to represent the current row of students and their seating arrangement.

The key idea is to use a 2D array dp where dp[i][mask] represents the maximum number of students that can be seated in row i with the given seating arrangement represented by the bitmask mask. We will iterate over each row and update the values of dp based on the previous row's values.

At each row i, we will iterate over all possible valid seating arrangements represented by the bitmask mask. For each valid mask, we will check if it satisfies the conditions and update the dp array accordingly.

Finally, the maximum number of students that can be seated on the bench will be the maximum value in the last row of the dp array. :

Solution Code

class Solution {
    public int maxStudents(char[][] seats) {
        int m = seats.length, n = seats[0].length;
        int[][] dp = new int[m + 1][1 << n];
        
        for (int i = 1; i <= m; i++) {
            for (int mask = 0; mask < (1 << n); mask++) {
                boolean valid = true;
                for (int j = 0; j < n; j++) {
                    if ((mask & (1 << j)) == 0) continue;
                    if (seats[i - 1][j] == '#') {
                        valid = false;
                        break;
                    }
                    if (j > 0 && ((mask & (1 << (j - 1))) != 0)) {
                        valid = false;
                        break;
                    }
                    if (j < n - 1 && ((mask & (1 << (j + 1))) != 0)) {
                        valid = false;
                        break;
                    }
                }
                
                if (valid) {
                    for (int prevMask = 0; prevMask < (1 << n); prevMask++) {
                        if ((prevMask & mask) == 0) {
                            dp[i][mask] = Math.max(dp[i][mask], dp[i - 1][prevMask] + Integer.bitCount(mask));
                        }
                    }
                }
            }
        }
        
        int maxStudents = 0;
        for (int mask = 0; mask < (1 << n); mask++) {
            maxStudents = Math.max(maxStudents, dp[m][mask]);
        }
        
        return maxStudents;
    }
}

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