LeetCode 1349: Maximum Students Taking Exam Solution

Master LeetCode problem 1349 (Maximum Students Taking Exam), a hard 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.

1349. Maximum Students Taking Exam

Hard

Array Dynamic Programming Bit Manipulation Matrix Bitmask

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming with bit masking. We iterate through each row of the classroom and for each row, we check all possible combinations of students sitting in the row without cheating. We use bit masking to represent the state of each row and store the maximum number of students that can be seated without cheating in a 2D dp array.

For each row, we iterate through all possible combinations of students and check if the current combination is valid (no cheating). If it is valid, we update the dp array with the maximum number of students that can be seated in that row.

Finally, we return the maximum value from the last row of the dp array as the result.

Time complexity: O(m * 2^n * 2^n), where m is the number of rows and n is the number of columns. Space complexity: O(m * 2^n), where m is the number of rows and n is the number of columns.

Solution Code

class Solution {
    public int maxStudents(char[][] seats) {
        int m = seats.length;
        int n = seats[0].length;
        
        int[][] dp = new int[m][1 << n];
        
        for (int i = 0; i < m; i++) {
            for (int mask = 0; mask < (1 << n); mask++) {
                if (isValid(seats, i, mask, n)) {
                    for (int prevMask = 0; prevMask < (1 << n); prevMask++) {
                        if ((prevMask & (mask << 1)) == 0 && ((prevMask << 1) & 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 - 1][mask]);
        }
        
        return maxStudents;
    }
    
    private boolean isValid(char[][] seats, int row, int mask, int n) {
        for (int i = 0; i < n; i++) {
            if ((mask & (1 << i)) != 0 && seats[row][i] == '#') {
                return false;
            }
        }
        return true;
    }
}

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